hi....
I want populate combo box based on another combobox selection in php.
for eg: If we select 'tamilnadu' in one combobox then tamilnadu district are displayed in another combobox.
If you know send that coding.
Thanks in advance
hi....
I want populate combo box based on another combobox selection in php.
for eg: If we select 'tamilnadu' in one combobox then tamilnadu district are displayed in another combobox.
If you know send that coding.
Thanks in advance
Hi frnds,
I need display mysql field as check box value.
<?php
mysql_connect("localhost","root","");
mysql_select_db("postal") or die(mysql_error());
$query = mysql_query("SELECT name FROM register");
while ($row = mysql_fetch_assoc($query)) {
$tag = $row['name'];
echo "<input type='checkbox' name='register[]' [U]value='$tag'[/U] />\n";
}
?>
here i paste my code here. it displayed only checkboxes. not value.any one know this send the corrections.
i frnds,
Many times i try to retrieve image from database. Successfully it upload to database. but cant retrieve. any one know this?
Thanks in advance....
Hi frnds,
I want the source codew for converting any one format to pdf. Any one know abt this pls help to me.
Thanks in advance.
Hi frnds,
I want activate the textbox when i select the particular checkbox in php.Anybody know send that coding ......
Hi Frnds,
I created session for index page. I passed session variable to next page like "about us" page. But in index page it displayed some errors.
Notice: Undefined index: myusername in C:\wamp\www\fc2\index.php on line 136
Notice: Undefined index: mypassword in C:\wamp\www\fc2\index.php on line 137
Line 136 : $myusername=$_POST;
Line 137 : $mypassword=$_POST;
this enough for find error or i paste my full program?
if you know help me pls.
Hi frnd,
Actually i write program for storing image path in database and retrieve the path then display image. if you know the code pls send me.
Hi frnd,
I used this coding. but it didnt display the value. now i paste my code and output here. Otherwise if you can give full coding for retrieving image from database.
seimg.php
<body>
<form action="seimg.php">
Enter image id:<input type="text" name="foo" />
<input type="submit" name="submit" value="show" />
<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "simg";
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
//check that $_POST['foo'] exists
if (isset($_POST['foo'])) {
//$_POST['foo'] exists
$foo = $_POST['foo']; //we might perform input validation here
} else {
//$_POST['foo'] does not exist
$foo = ''; //assign a default value
}
//use $foo
echo $foo;
$query = mysql_query("SELECT * FROM img WHERE id='".$foo."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header('Content-type: image/jpg');
echo $content;
?>
</form>
</body>
</html>
Output:
[url]http://localhost/base/seimg.php?foo=1&submit=show[/url]
the same error:
"Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24"
Hi frnd,
really thanks for reply. But again it shows an error. Can you help me for this?
actually error pointing
$b =$_GET; this line only. i dont know wat error in this line. Any one know say to me.
Hi frnds,
I did same wat u said. but it shows an error. I dont know. if u can pls say to me. I paste my code here.
seimg.php
<body>
<form action="seimg.php">
Enter image id:<input type="text" name="field" />
<input type="submit" name="submit" value="show" />
<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "simg";
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
$b =$_GET['field'];
$query = mysql_query("SELECT image FROM img WHERE id='$b'");
$row = mysql_fetch_array($query);
echo "<img src='".$row['image']."' />";
?>
</form>
it shows "Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24" this error.
I used this. But it shows an error."Parse error: parse error in C:\wamp\www\base\seimg.php on line 30"
line 30:echo "<img src='". <?php echo $row?>."' />";
Hi frnds,
I used $b =$_REQUEST; It shows error in this line."Notice: Undefined index: field in C:\wamp\www\base\seimg.php on line 24" .
Hi..
1.
$query = mysql_query("SELECT * FROM img WHERE id='$b'");
2.
3.
$row = mysql_fetch_array($query);
4.
echo "<img src='".$content = $row."' />";
this coding working.But it not displayed image. It displayed some binary data. wat error in this ?
hi frnd,
I use that link and do the same way. but i dont know how link one file to another in this program . successfully image downloaded, but not stored into database. can you help me?
Hi frnds,
I want to store images in database and retrieve. I can successfully stored. But i dont know how retrieve it. If any one know pls help me. I paste my code here.
fimg.php
<body>
<form enctype="multipart/form-data" action="insert.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>
</body>
insert.php
<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "simg";
$link = mysql_connect($host, $username, $password);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db ($database);
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
$tmpName = $_FILES['image']['tmp_name'];
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
$query = "INSERT INTO img";
$query .= "(image) VALUES ('$data')";
$results = mysql_query($query, $link);
print "Thank you, your file has been uploaded.";
}
else {
print "No image selected/uploaded";
}
mysql_close($link);
?>
seimg.php
<form action="seimg.php">
Enter image id:<input type="text" name="field" />
<input type="submit" name="submit" value="show" />
<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "simg";
@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
@mysql_select_db($database) or die("Can not select the database: ".mysql_error());
$b =$_REQUEST['field'];
$query = mysql_query("SELECT * FROM img WHERE id='$b'");
$row = mysql_fetch_array($query);
$content = $row['image'];
?>
Really thanks urtrivedi and rajarajan. Now am working on it.
I dont know how checked in phpmyadmin. Bcoz i write query like "$idvalue=$_REQUEST". So according user selection it will retrieve the values. I don't know how check it in phpmyadmin.
1) I need "where" condition for the first query only for retrieving values from table.
I passed id from php . This id passed to next detail.php file like $idvalue=$_REQUEST; Then retrieve value from old table using where id='$idvalue.This underline id is one field of old table.
Ya echo "$idvalue" display the value. just I want retrieve some values from existing table and posted into new table. I used where condition here for retrieve values. But no condition for storing values.already i done this in following way. select from one table and printed in one place and then moved to another table. I thing this long process. so only i need this way. sorry frnd my query also too long.
Hi thanks for your reply.......
Just I insert your coding in phpmy admin it produce the following error.
Error
SQL query:
$query = "insert into orde values('$projectid','$projectname','$projectcat','$scope','$location','$rate')";
MySQL said: Documentation
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query="insert into orde values('$projectid','$projectname','$projectcat','$scop' at line 1
And I add some codings with this then i show some error
"select * from smenu where id='$idvalue'";
while($row=mysql_fetch_array($result))
{
$projectid=$row['id'];
$projectname=$row['main'];
$projectcat= $row['itemm'];
$scope=$row['descrip'];
$location=$row['Category'];
$rate=$row['Rate'];
}
$query="insert into orde values('$projectid','$projectname','$projectcat','$scope','$location','$rate')";
$dbase=mysql_query($query);
echo $dbase;
error_reporting(E_ALL);
ini_set("display_errors", 1);
error
Error
Error
SQL query: Documentation
"select * from smenu where id='$idvalue'";
MySQL said: Documentation
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"select * from smenu where id='$idvalue'"' at line 1
echo statement not necessary here. just i know , items are displayed are not.I used ur query. But still didn't store values in table.
All items are displayed using echo statement. but not stored in table . not shown any error.
Hi frnds,
I need to select some values in old table and send it into newtable. Here i placed my program. Anybody know say the corrections.
<?php
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$idvalue=$_REQUEST['id'];
//$iv=$_REQUEST['z'];
$result=mysql_query("select * from smenu where id='$idvalue'");
while($row=mysql_fetch_array($result))
{
$projectid=$row['id'];
$projectname=$row['main'];
$projectcat= $row['itemm'];
$scope=$row['descrip'];
$location=$row['Category'];
$rate=$row['Rate'];
}
echo " $projectid,$projectname,$projectcat,$scope,$location,$rate";
$query="insert into orde values('$projectid','$projectname','$projectcat','$scope','$location','$rate' where id='$idvalue')";
$dbase=mysql_query($query);
echo "$dbase";
?>
</body>
</html>
$sq="Select SUM(total) from bill where invoiceno=$q";
$r=mysql_query($sq);
echo "</table>";
echo "Total";
echo $r;
I try this coding. But it shows me an output "TotalResource id #5".But the original answer is 39. any error this coding? Ask me frnd..
now i try this frnd. but one more doubt. Now i clearly say.
insert into table2 , select * from table1 ( It table1 have some empty field). I try to do this Retrieve value from one table and stored into another table. I dont know how do it nu......
Really very very thanks of this...
Hi frnds,
I need to perform calculation in same field .
$sql="SELECT itemcode,itemname,quantity,price,total,remarks,SUM(total) FROM bill WHERE invoiceno = '".$q."'";
Is this query possible?anybody know say to me pls.
Add Item.php
<body>
<script type="text/javascript">
function showUser(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getite.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form action="getite.php">
<h3><a href="getitem.php">Add Item</a></h3>
<h3><a href="http://localhost/da/logi.php">Home</a></h3><br/>
<?php
error_reporting(0);
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$query="SELECT DISTINCT main FROM smenu";
$result = mysql_query ($query);
echo "<select name='sers' onchange='showUser(this.value)'>";
// printing the list box select command
while($nt=mysql_fetch_array($result)){
echo "<option value='$nt[main]'>$nt[main]</option>";
}
echo "</select>";
?>
<div id="txtHint"><b>Items info will be listed here.</b></div>
</form>
</body>
</html>
getite.php
<body>
<?php
$q=$_GET["q"];
echo "$q";
error_reporting(0);
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$sql="SELECT * FROM smenu WHERE main = '".$q."'";
$result = mysql_query($sql);
mysql_close($con);
echo "<table width=900 border=1>";
echo "<tr bgcolor=#ccccc>";
echo "<th width=80 scope=col>id</th>";
echo " <th width=100 height=34 scope=col>Item</th>";
echo "<th width=500 scope=col><div align=center>Description</div></th>";
echo "<th width=300 scope=col><div align=center>Category</div></th>";
echo "<th width=94 scope=col>Rate</th>";
echo "<th width=94 scope=col>Order Item</th>";
echo "</tr>";
while($row=mysql_fetch_array($result))
{
echo "<tr><td align='center'>";
echo $row['id'];
echo "</td><td align='center'>";
echo $row['itemm'];
echo "</td><td align='left'>";
$chec=$row['descrip'];
if ($chec == ' ' or $chec == Null)
echo 'NILL';
else
echo $row['descrip'];
echo "</td><td align='center'>";
$check = $row['Category'];
if ($check == ' ' or $check == Null)
echo 'NILL';
else
echo $row['Category'];
echo "</td><td align='center'>";
echo $row['Rate'];
echo "</td><td align='center'>";
echo "<a href=detail.php?id=".$row['id'].">Order</a>";
echo "</td></tr>";
}
echo "</table>";
?>
</body>
</html>
detail.php
<form id="form1" name="form1" method="post" action="accep.php">
<?php
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$idvalue=$_REQUEST['id']; …
Really very thanks frnds. I got error when I insert empty field into table.
How we avoid this error. If I enter something empty field it displays following error.
Notice: Undefined index: ploca in C:\wamp\www\data\accep.php on line 20
Notice: Undefined index: psco in C:\wamp\www\data\accep.php on line 19
Hi ..........
I attached screen shot with this thread. is this possible for calculating these two values and display the same page. This values retrieve from database.
Really thanks . I successfully storing values.
I got this error message
Notice: Undefined index: plo in C:\wamp\www\data\accep.php on line 19
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order values('1','starters','Edamame','Steamed','','4.95')' at line 1
Notice: Undefined index: plo in C:\wamp\www\data\accep.php on line 19
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order values('1','starters','Edamame','Steamed','','4.95')' at line 1
Hi friends,
I get some error for storing into database. I given that coding.any body know pls say to me.
accep.php
<?php
$check=$_REQUEST['submit'];
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$projectid=$_REQUEST['pid'];
$projectname= $_REQUEST['pname'];
$projectcat= $_REQUEST['pcat'];
$scope=$_REQUEST['psco'];
$location=$_REQUEST['plo'];
$rate=$_REQUEST['prate'];
$query="insert into order values('$projectid','$projectname','$projectcat','$scope','$location','$rate')";
$dbase=mysql_query($query) or die(mysql_error());
if($dbase==TRUE)
{
include("congrat.php");
}
?>
Detail.php
<body>
<h3> <a href="additem.php">View Menu</a></h3>
<form id="form1" name="form1" method="post" action="accep.php">
<?php
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$idvalue=$_REQUEST['id'];
$result=mysql_query("select * from smenu where id='$idvalue'");
while($row=mysql_fetch_array($result))
{
$projectid=$row['id'];
$projectname= $row['main'];
$projectcat= $row['itemm'];
$scope=$row['descrip'];
$location=$row['Category'];
$rate=$row['Rate'];
}
echo "<input type='hidden' value=".$projectid." name=pid>";
echo "<input type='hidden' value=".$projectname." name=pname>";
echo "<input type='hidden' value=".$projectcat." name=pcat>";
echo "<input type='hidden' value=".$scope." name=psco>";
echo "<input type='hidden' value=".$location." name=plo>";
echo "<input type='hidden' value=".$rate." name=prate>";
/*
echo " project name is $projectname";
if($check==="Request for this")
{
echo " project id is $projectid <br>";
echo "check value is $check";
}*/
?>
<table width="380" height="363" border="0" align="center" cellpadding="4" cellspacing="3" >
<tr>
<td width="135">Item Id </td>
<td width="211"><?php echo $projectid; ?></td>
</tr>
<tr>
<td>Main Item </td>
<td><?php echo $projectname; ?></td>
</tr>
<tr>
<td>Item Name</td>
<td><?php echo $projectcat; ?></td>
</tr>
<tr>
<td height="83">Description</td>
<td><?php echo $scope; ?></td>
</tr>
<tr>
<td height="83">Category</td>
<td><?php echo $location; ?></td>
</tr>
<tr>
<td>Rate</td>
<td><?php echo $rate; ?></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td><input name="submit" type="submit" id="submit" value="Order this" /></td>
</tr>
</table>
additem.php
<script type="text/javascript">
function showUser(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getite.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form action="getite.php">
<h3><a href="getitem.php">Add Item</a></h3><br/>
<?php
error_reporting(0);
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$query="SELECT DISTINCT main FROM smenu";
$result = mysql_query ($query);
echo "<select name='sers' onchange='showUser(this.value)'>";
// printing the list box select command
while($nt=mysql_fetch_array($result)){
echo "<option value='$nt[main]'>$nt[main]</option>";
}
echo "</select>";
?>
<div id="txtHint"><b>Items info will be listed here.</b></div>
</form>
</body>
</html>
getite.php
<?php
$q=$_GET["q"];
echo "$q";
error_reporting(0);
mysql_connect("localhost","root","");
mysql_select_db("menus") or die(mysql_error());
$sql="SELECT * FROM smenu WHERE main = '".$q."'";
$result = mysql_query($sql);
mysql_close($con);
echo "<table width=800 border=1 cellspacing=2>";
echo "<tr bgcolor=#ccccc>";
echo " <th width=100 height=34 scope=col>Item</th>";
echo "<th width=500 scope=col><div align=center>Description</div></th>";
echo "<th width=300 scope=col><div align=center>Category</div></th>";
echo "<th width=94 scope=col>Rate</th>";
echo "<th width=94 scope=col>Order Item</th>";
echo "</tr>";
while($row=mysql_fetch_array($result))
{
echo "<tr><td align='center'>";
echo $row['itemm'];
echo "</td><td align='left'>";
echo $row['descrip'];
echo "</td><td align='center'>";
echo $row['Category'];
echo "</td><td align='center'>";
echo $row['Rate'];
echo "</td><td align='center'>";
echo "order this";
}
echo "</table>";
?>
</body>
</html>
If Click "order this", then the particular item goes to separate database. How to do this..
i cant attached document in existing thread. there is no option. "Insert Link" only available.
I attached screen shot with this thread. how we display data from database without null values.
Really thanks for quick reply. I attach one more screen shot. I used list and select the value it retrieve data from database. some fields only empty. You see attachment you get clear idea of my question.
hi........
I want display data from database. I done it. but some fields have null values. it display empty field . I attached screen shot with this. If you know the solution pls say to me.
hi friends,
I used following code for automatically incrementing value and stored in database. This working but it have small problem. when I submit value then only incrementing. ex, i entered details for invoice number 2, but it showing invoice number 1.
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("menus") or die(mysql_error());
$query2=mysql_query("select invoiceno from bill");
while($row=mysql_fetch_array($query2))
{
$var=$row['invoiceno'];
}
$count=$var+1;
echo $count;
?>
anybody knowing result for pls help to me.
i want select values in list ,then it automatically stored in database. how to do ?
if anybody know say to me....
ya i try this. its working. i have one more doubt. values from list to text box successfully working. using this value can we retrieve data from database and displayed in another text box. u cant understand na i will send screen shot.
Thanks...
hi friends,
i want calculating values in two text box , result display in another text box and this result will be stored in database. if anybody know this say to me..