simple search question from query.

[jbrill]

I am trying to use a form to call another page (search.php) where search.php finds the text enter in the text field from the database and displays it.

The problem im having is that I cannot seem to get it working... could some one please look at  my code and tell me if it is ok, and if it isn't, help me make it work.

 

My error Im getting is: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/morow/public_html/admin/search.php on line 68

 

but i think it has to do with the way i query the database....im new to php...

 

Help is much appreciated, this forum is awesome!

 

 

\\assuming I have already connected tot he database

<?
echo"<table border=\"0\" width=\"80%\" bgcolor=\"#ffffff\" class=\"MainBody1\" cellpaddin=\"0\" cellspacing=\"0\">";
echo"<tr>";
echo"<td class=\"underline\">Quote Number</td><td class=\"underline\">Job Number</td><td class=\"underline\">PO Number</td><td class=\"underline\">Customer</td><td class=\"underline\">Status</td><td class=\"underline\">Action</td>";
echo"<tr>";


$result = mysql_query("SELECT * FROM jobs WHERE prod_number='".$_GET['prod_number']."'");

while($row = mysql_fetch_array($result))
{

	 echo "<tr><td>" . $row['id'] .  "</td><td>"  . $row['job_number'] ."</td><td>" . $row['po_number'] . "</td><td>" . $row['customer'] . "</td><td>" . $row['status'] . "</td><td> 
  
  
  <a class=\"link\" href=\"admin_modnews.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionEdit.gif\" border=\"0\"></a> <a class=\"link\" href=\"admin_delete.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionDelete.gif\" border=\"0\"></a>
  
  
  </td></tr>";

}
echo"</table>";
?>

[redarrow]

try this please.

 

<?php
echo"<table border=\"0\" width=\"80%\" bgcolor=\"#ffffff\" class=\"MainBody1\" cellpaddin=\"0\" cellspacing=\"0\">";
echo"<tr>";
echo"<td class=\"underline\">Quote Number</td><td class=\"underline\">Job Number</td><td class=\"underline\">PO Number</td><td class=\"underline\">Customer</td><td class=\"underline\">Status</td><td class=\"underline\">Action</td>";
echo"<tr>";


$result = mysql_query("SELECT * FROM jobs WHERE  like prod_number='%".$_GET['prod_number']."%' ");

while($row = mysql_fetch_array($result))
{

	echo "<tr><td>" . $row['id'] .  "</td><td>"  . $row['job_number'] ."</td><td>" . $row['po_number'] . "</td><td>" . $row['customer'] . "</td><td>" . $row['status'] . "</td><td> 
  
  
  <a class=\"link\" href=\"admin_modnews.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionEdit.gif\" border=\"0\"></a> <a class=\"link\" href=\"admin_delete.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionDelete.gif\" border=\"0\"></a>
  
  
  </td></tr>";

}
echo"</table>";
?>

[jbrill]

still no luck....

 

got this warning: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/morow/public_html/admin/search.php on line 68

 

line 68 is:

 

while($row = mysql_fetch_array($result))

[redarrow]

<?php
echo"<table border=\"0\" width=\"80%\" bgcolor=\"#ffffff\" class=\"MainBody1\" cellpaddin=\"0\" cellspacing=\"0\">";
echo"<tr>";
echo"<td class=\"underline\">Quote Number</td><td class=\"underline\">Job Number</td><td class=\"underline\">PO Number</td><td class=\"underline\">Customer</td><td class=\"underline\">Status</td><td class=\"underline\">Action</td>";
echo"<tr>";


$result = mysql_query("SELECT * FROM jobs WHERE  prod_number like ='%".$_GET['prod_number']."%' ");

while($row = mysql_fetch_array($result))
{

	echo "<tr><td>" . $row['id'] .  "</td><td>"  . $row['job_number'] ."</td><td>" . $row['po_number'] . "</td><td>" . $row['customer'] . "</td><td>" . $row['status'] . "</td><td> 
  
  
  <a class=\"link\" href=\"admin_modnews.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionEdit.gif\" border=\"0\"></a> <a class=\"link\" href=\"admin_delete.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionDelete.gif\" border=\"0\"></a>
  
  
  </td></tr>";

}
echo"</table>";
?>

[jbrill]

still no luck... same error

[redarrow]

post the select statement when tried this ok.

 

<?php
echo"<table border=\"0\" width=\"80%\" bgcolor=\"#ffffff\" class=\"MainBody1\" cellpaddin=\"0\" cellspacing=\"0\">";
echo"<tr>";
echo"<td class=\"underline\">Quote Number</td><td class=\"underline\">Job Number</td><td class=\"underline\">PO Number</td><td class=\"underline\">Customer</td><td class=\"underline\">Status</td><td class=\"underline\">Action</td>";
echo"<tr>";

$x=$_GET['prod_number'];
$result = " SELECT * FROM `jobs` WHERE `prod_number` like ='%$x%' ";
$result=mysql_query($result)or die("mysql_error");


echo $result;

while($row = mysql_fetch_array($result))
{

	echo "<tr><td>" . $row['id'] .  "</td><td>"  . $row['job_number'] ."</td><td>" . $row['po_number'] . "</td><td>" . $row['customer'] . "</td><td>" . $row['status'] . "</td><td> 
  
  
  <a class=\"link\" href=\"admin_modnews.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionEdit.gif\" border=\"0\"></a> <a class=\"link\" href=\"admin_delete.php?idr=".$row['id']."&table=jobs\"><img src=\"images/actionDelete.gif\" border=\"0\"></a>
  
  
  </td></tr>";

}
echo"</table>";
?>

[jbrill]

still no luck! getting an error in the query still and its the correct way...

[teng84]

is this declaredas char or varchar in your db "prod_number`"

 

the way you wrote it seems like a number number i guess cant use like