Hello, i would like to know if its any chance of me being able to retrive data by typing ID, so for example if i type 100 and click on get, i want it to echo the "mobile" number. Can someone please help me with that
Thank you
NoID
0
Light Poster
Recommended Answers
Jump to PostSounds straightforward. You're using a form I take it with a database full of mobile numbers, correct?
Create your form with a text box, give it a name attribute of 'myId' or something.
Your php form handling script will then pick up this data like so:
$myIdToCheck = …
Jump to PostYou haven't said what $result is. You can't ask a function to return the contents of the $result object if it doesn't exist.
DO this:
[B]$result = [/B]mysql_query("SELECT ID FROM table WHERE ID = '".$_POST['ID']."'LIMIT 1");I still don't understand your sql though, why are you …
Jump to Postmake sure that your query is properly present inside the if loop.
It can be like this:
if(isset($_POST['ID'])){ mysql_query("SELECT ID FROM table WHERE ID = '".$ID."'LIMIT 1"); $row = mysql_fetch_assoc($result); echo $row; mysql_close($con); }
All 10 Replies
diafol
Sounds straightforward. You're using a form I take it with a database full of mobile numbers, correct?
Create your form with a text box, give it a name attribute of 'myId' or something.
Your php form handling script will then pick up this data like so: $myIdToCheck = $_POST['myId']; check your database for the id: SELECT * FROM mytable WHERE id = '{$myIdToCheck}' The above is extremely lazy and doesn't include security or validation features. Read some online tutorials on $_POST, form submission, SQL statements and recordset manipulation in php. You'll need to read up on these before using this in a live website.
NoID
0
Light Poster
<html>
<head>
</head>
<body>
<?php
$con = mysql_connect("localhost","t","123456");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("t", $con);
if(isset($_POST['ID'])){
mysql_query("SELECT ID FROM table WHERE ID = '".$ID."'LIMIT 1");
}
$row = mysql_fetch_assoc($result);
echo $row;
mysql_close($con);
?>
<form method="post" action="">
<font>ID</font> <input type="text" name="ID"/>
<input type="Submit" value="Submit" name="Submit">
</form>
</body>
</html>I have got this but i am getting an error
Notice: Undefined variable: result in C:\wamp\www\test.php on line 17
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\wamp\www\test.php on line 17
any help please
guru12
-2
Light Poster
Dear friend
Assign the $ID
$ID=$_POST;
Use the below code
<html>
<head>
</head>
<body>
<?php
$con = mysql_connect("localhost","t","123456");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("t", $con);
if(isset($_POST['ID'])){
mysql_query("SELECT ID FROM table WHERE ID = '".$_POST['ID']."'LIMIT 1");
}
$row = mysql_fetch_assoc($result);
echo $row;
mysql_close($con);
?>
<form method="post" action="">
<font>ID</font> <input type="text" name="ID"/>
<input type="Submit" value="Submit" name="Submit">
</form>
</body>
</html>Thanks and Regards
NoID
0
Light Poster
Hello Friend,
after trying what you said i am still getting this error
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\wamp\www\test.php on line 17
What i want to do is, just be able to type ID, and it echos the email of that ID.
Thank you
diafol
You haven't said what $result is. You can't ask a function to return the contents of the $result object if it doesn't exist.
DO this:
[B]$result = [/B]mysql_query("SELECT ID FROM table WHERE ID = '".$_POST['ID']."'LIMIT 1");I still don't understand your sql though, why are you retrieving just the ID when you know it already from the post variable??
praghna
0
Newbie Poster
make sure that your query is properly present inside the if loop.
It can be like this:
if(isset($_POST['ID'])){
mysql_query("SELECT ID FROM table WHERE ID = '".$ID."'LIMIT 1");
$row = mysql_fetch_assoc($result);
echo $row;
mysql_close($con);
}
diafol
make sure that your query is properly present inside the if loop.
It can be like this:
[code:
if(isset($_POST)){
mysql_query("SELECT ID FROM table WHERE ID = '".$ID."'LIMIT 1");
$row = mysql_fetch_assoc($result);
echo $row;
mysql_close($con);
}
/code]
ALthough better, that still leaves out declaring the $result object. So it will not work.
NoID
0
Light Poster
After trying the following code, i can confirm that i dont get any errors, but when i insert the ID it comes back and says "Array", i really want to retrive the email, but i just want to see if i am able to retrieve the ID back or not.
diafol
If you echo $row , you'll get an array because it's returning a whole row of the resultset, regardless of the fact that there's only one field.
You need to be explicit:
echo $row['id'];
NoID
0
Light Poster
That works like a dream, the reason why i was trying to do this, was so i can see how to retrieve data, so i can enter the ID and it prints email, but how am i able to send an email and not just echo an email.
$to = 'nobody@example.com';
$subject = 'the subject';
$message = 'hello';
$headers = 'From: webmaster@example.com' . "\r\n" .
'Reply-To: webmaster@example.com' . "\r\n" .
'X-Mailer: PHP/' . phpversion();
mail($to, $subject, $message, $headers);I have been trying to mess around with the code that you all guys have helped but unfortunately i haven’t been successful.
Thank you for your help.
Have a good day.
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