Posts by nav33n

Why not use order by clause in your query itself ?

$query = "select * from table order by colname [asc | desc ]";

[asc | desc ] is optional. You can sort the records in ascending order or descending order on one or multiple columns.
Or, you can also fetch the records from the table into an array and sort the array and get the max or min value from it.

Great ! Congrats :)

Yep. An array of checkboxes. When the user submits the form, only those checkboxes which were selected, will be posted. You can then build your query depending on the checkboxes the user selects. :)

I don't know whats causing the error. You can run queries in a for loop. Once you have freed the result using mysql_free_result, that resource identifier is freed. Actually, if your query isn't too complicated, you don't have to use free_result as php frees the result at the end of execution of the script. Source: http://nl3.php.net/mysql_free_result

Can you post your code here ?

Yeah.. You can store the description in whatever way you want..

Yes. Using insert query you can insert anything, long text, short text, image, etc :)

What do you mean by describe it ?

You are welcome.

No. You can't store a value in the table without using an insert query.

Y/w. I ll be here.. ( I hope!)

When you execute a query, a resource identifier will be assigned to $result. mysql_free_result will free this value.

$q = "show tables";

I have 26 tables...shall i write all the names at once

$q="show a,b,c...z";

Show tables will get all the 26 tables (If you have 26 tables in the database) . Thats a mysql query.

No. The code that I have given will search for all the tables in the database. (make sure you have only those tables in the database which you want to search). And what I mean by print your query is, echo $searchtables;
Tell us what it prints on the screen.

Why ? print_r($arrayname) should work fine.

Have you changed the column and table names to suit your requirements ? Print out $searchtables and tell us what it prints.

You have to re-write your code. Or, you can search the internet for a software (if there is any) to convert jsp application to php.

<?php
$var = array( ); //declare an array
$query = "select * from table";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)) {
  $var[] = $row['name']; //assign a value 
}
?>

Note: The code works even without declaring a variable as array.
But, you can't use a normal variable to act like an array. Example,

<?php
$x = 100;
$x[] = "10";
print $x;
?>

Print the query and show us what it says..

Basically, you need another table (comments) with columns to store who wrote the comment, on whose profile was the comment written and the comment. (You also may need more columns like date). I hope every user has a unique userid. Store the userids ("from" and "to" ) and comment in the table. Then while displaying the profile, check if there are any entries for this userid. If any rows found, display it.

As you have already said it yourself, you don't have to explicitly declare a variable as an array to use it like an array. You can simply use $var[] = "value"; This will treat $var as an array. But, its always a good practice to initialize a variable before using it. (But most of them don't do it ;) )

Umm.. Is this how you want your script to work ? The user can select different checkboxes and enter a search string next to the checkbox. Then when he clicks on search, It should return all the matching records ?

You can do it this way.

<?php
$con = mysql_connect("localhost","root");
mysql_select_db("test");
//get all the tables from the database test
$q = "show tables";
$r = mysql_query($q);
$searchresult = array();
while($row = mysql_fetch_array($r)) {
	$table = $row[0]; // get the tablename
	$searchtables = "select * from ".$table." where col1 like '%".$searchstring."%'"; //search the table column for the search string
	$result = mysql_query($searchtables);
	while($rows = mysql_fetch_array($result)) {
		$searchresult[] = $rows['name'];
	}
}
// $searchresult will have all the records from all the tables for the searchstring
?>

Cheers,
Nav

Awesome!

Yep.. My example works with 2 dropdowns in a same page.

:) lol.. no ! I am a learner.. I am glad your problem was fixed !

What second page ? Why dont you keep everything in the same page ? :-/

I'm in a situation where I have to pass javascript variable to php without redirect (no $_get[]).

OP said he don't want to use $_GET[] ! I think he wants something like this.

<script type='text/javascript'>
var xyz;
xyz = 1000;
<?php 
$value = xyz; // he wants xyz value of javascript to be assigned to php variable  $value. 
?>
</script>

//But this is possible
<?php
$value = 100;
?>
<script type='text/javascript'>
var xyz;
xyz = <?php echo $value; ?>;
alert (xyz);
</script>

AFAIK, you can do it the other way, ie., assigning a php variable value to a javascript variable. Assigning a javascript variable value to a php variable is not possible since php is a server side scripting language.

Use firefox to execute your scripts. Make use of error console to see any javascript error.

I don't know what you mean by that.. But, if you use "if file exists" and add a random number to the image, What is the point in storing the filename in the table ?

Umm.. I haven't used crons.. But this can be of some use to you on how to run crons. http://www.modwest.com/help/kb5-125.html
Note: You need linux to run cron jobs.

is it possible to a random number to a image file without it becoming a part of that file name.

No.. You can rename a file by concating a random number. You cant add random number to an image without changing its name..

You can do it in 2 ways. Configure a cron job to run everyday at a certain time and send out mails if there are any records for that particular date. The other way is to write a script and execute it manually everyday.

Then use a sub query.

$query = "select D_Name from Department where col_No = (select col_No from College where col_name = '$firstdropdownlistvalue')";

This will return the D_name of the selected col_name.

Ah ! Then, *I think* you need a loop to do that.

If you know the values, you can do something like this.

insert into table (col1) values (val1),(val2),(val3)

This will insert 3 records to the table.

What do you mean by without using for or while loop ? What's the problem in inserting a record directly ? ie., insert into tablename (col1,col2) values ('val1','val2')

onchange="javascript: document.delete.submit();" instead of
onchange=javascript: document.delete.submit();

Secondly,

$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$get_list_result2'");

is wrong. What are you trying to do ?

The above query doesn't work because, it will search for the record where shape is like "$searchshapeone" AND again, like, "$searchshapetwo". It will work only if there is a record in the database with shape like both $searchshapeone and $searchshapetwo. For example, If there is a record in the table with shape, for example, 12345, if $searchshapeone = 123 and $searchshapetwo = 345, then your query will work (because both conditions ie., shape like '%123%' and shape like '%345%' are satisfied). Try out with OR condition. It will work !

Post your latest code.

I said it in the other thread, have an onchange event to submit the page when you select a value from the dropdown list. When the page is submitted, you can know which option was selected from the dropdown list. ($val = $_POST; ) Then query the table with this value as condition and retrieve the data.

Hi, the above code works fine. You just have to change the tablenames/column names. Anyway, When the page gets submitted, echo the value of 1st dropdownlist. See if the value is passed correctly. If yes, then the problem must be with your 2nd query. Check your query. See if your query is right. If it is correct, then, something is going wrong while fetching the records from the table. :) You should check this yourself because I dont have the access to your table and I don't know what exactly is the problem.

Okay ! Well, I just re-wrote the whole code and it works ! Change the column names and tablename to suit your needs ! :) If this doesn't work, then, I don't know, something must be wrong from your side.

<?php
	$con = mysql_connect("localhost","root");
	mysql_select_db("test");
	
	$firstdropdownlistvalue = $_POST['dropdown1'];
	
	$query = "select order_id from orders";
	$result = mysql_query($query);
	echo "<form method='post' name='form1' action='test.php'>";
	echo "<select name=dropdown1 onchange='javascript: document.form1.submit();'>";
	while($row = mysql_fetch_array($result)) {
		$value = $row['order_id'];
		if($value == $firstdropdownlistvalue) { $selected = "selected"; } else { $selected = ""; }
		echo "<option value='".$value."' $selected>".$value."</option>";
	}
	echo "</select>";
	
	$query2 = "select * from orders where order_id = '$firstdropdownlistvalue'";
	$result2 = mysql_query($query2);
	echo "<select name=dropdown2>";
	while($row2 = mysql_fetch_array($result2)) {
		$value2 = $row2['amount'];
		echo "<option value='".$value2."'>$value2</option>";
	}
	echo "</select>";
	echo "</form>";
?>

Hey, If you don't have any "processing" in your script, why not redirect the script to the same page ? Will it still have the images in the cache ?

First of all, you don't have an onchange event to submit the page. secondly, you should have a value for the option. Here is a simple example. I haven't tested it, but it should work with little tuning (which you have to figure it yourself !)

<?php
	$con = mysql_connect("localhost","root");
	mysql_select_db("test");
	$query = "select col1 from table";
	$result = mysql_query($query);
	echo "<form method='post' action='test.php'>";
	echo "<select name=dropdown1 onchange='javascript: document.form.submit();'>";
	while($row = mysql_fetch_array($result)) {
		$value = $row['col1'];
		echo "<option value='".$value."'>$value</option>";
	}
	echo "</select>";
	
	$firstdropdownlistvalue = $_POST['dropdown1'];
	$query2 = "select * from table where col2 = '$firstdropdownlistvalue'";
	$result2 = mysql_query($query2);
	echo "<select name=dropdown2>";
	while($row2 = mysql_fetch_array($result2)) {
		$value2 = $row2['col1'];
		echo "<option value='".$value2."'>$value2</option>";
	}
	echo "</select>";
	echo "</form>";
?>

You can find anything on google ! Well, Just install eclipse and install this plugin..
http://sourceforge.net/project/showfiles.php?group_id=57621

Yep ! <head> tag is the right place to put meta tags. I haven't come across caching problem, but maybe you can use this.

<?php
header("Cache-Control: no-cache, must-revalidate"); // HTTP/1.1
header("Expires: Mon, 26 Jul 1997 05:00:00 GMT"); // Date in the past
?>

Source: http://nl2.php.net/header Ex. 2

Can you post your code here ?

heh.. no probs ! As the warning itself says, chmod didn't find the file (and hence unlink didn't work). Check if the path to the file is correct for chmod and unlink!

:) remove @ and check if it throws any error. You are supposed to delete the file with the extension !

Check if the number is greater than 0. If it is, then show it in green and negative in red. :-/ What is the problem ?

This has been discussed so many times. Either use ajax or this way. In the first dropdown list, list all the states. Have an onchange event to submit the form for the 1st dropdown list. When a user selects an option from the 1st dropdown list, onchange event will be fired and the form will be submitted. You can then access the selected value of 1st dropdown list by doing $firstdropdownlistvalue = $_POST['dropdownlistname']; You can then query the database for this particular state and get all the cities in the 2nd dropdown list.