Table display error

PHP
1

Hello All,

I have made a page with a search bar. It searches for keywords from three columns in a table. Now the problem I am facing is, I want the table to appear after I search for a keyword. But what is happening is, the header of the table, is coming before the submission itself. And along with it, I get a notice and warning.

Notice: Undefined variable: result in C:\xampp\htdocs\sdis\data\search-2d-data.php on line 35
Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\sdis\data\search-2d-data.php on line 35

Here is my code, can someone please help me.


<?php
  include $_SERVER["DOCUMENT_ROOT"].'/sdis/core/init.php';
  include $_SERVER["DOCUMENT_ROOT"].'/sdis/includes/overall/header.php';
  if(isset($_POST['search'])===true){
  	$search=$_POST['search'];
$query = "SELECT * FROM `2d_raw_data` WHERE `basin` LIKE '%".$search."%' OR `area` LIKE '%".$search."%' OR `block` LIKE '%".$search."%'";
  $result=mysql_query($query);
}
?>

<form id="form1" name="form1" method="post" action="">
<h2><center>Search 2D Data</center></h2>
<input type="text" class="form-control" name="search" required/><br/>
<center><button class="btn btn-lg btn-primary" type="submit" value="Search">Search</button></center><br/>
</form><br/><br/>


<table width="100%" border="1" cellspacing="1" cellpadding="0">
<tr>
<td>
<table width="100%" border="1" cellspacing="1" cellpadding="3">
<tr>
<td colspan="50"><strong><center>2D Data</center></strong> </td>
</tr>

<tr>
<td align="center"><strong>Basin</strong></td>
<td align="center"><strong>Area</strong></td>
<td align="center"><strong>Block</strong></td>


</tr>

<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><?php echo $rows['basin']; ?></td>
<td><?php echo $rows['area']; ?></td>
<td><?php echo $rows['block']; ?></td>
</tr>

<?php
}
?>

</table>
</td>
</tr>
</table>

<?php include $_SERVER["DOCUMENT_ROOT"].'/sdis/includes/overall/footer.php'; ?>
2

I think that means the query is returning zero results.

Also you should not be using the mysql_* functions as they are depreciated. More information http://www.php.net/manual/en/intro.mysql.php

3

Yes, the query isn’t returning anything useful for some reason.

I’m assuming the connection code is in the included (maybe require would be better here?) “init.php” file?

You could do 2 things that should help you troubleshoot.
You could echo out the query to make sure it looks like what you expect it to.
You can add an “or die mysql_error()” to the line that runs the query.

  • though using mysqli would be better - you should change over now while working on the code IMHO
4

If you are saying that when the page is initially loaded up the table header is appearing, then var_dump($_POST[‘search’]) and see what is causing that condition to fail.


  if(isset($_POST['search'])===true)

BTW, as that is checking if the anything has been typed in the search box, why not test for something a bit more useful.


  if(strlen($_POST['search']) > 2)

At least here you are getting 3 chars and your search has got a bit of a chance of working …

The error is caused because you have no conditional (if) check where you loop through any potential results - so it is trying to loop through a non-existent result set.