Problems w/ mysql_fetch_array()

[Jaume_hernandez]

Hi all!!!I've adapted the code of http://www.w3schools.com/php/php_ajax_livesearch.aspto my MySQL table so I can show a live search with AJAX. Evething looks fine but when I choose the category a get an error mesage:Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\UsedMachines\getuser.php on line 25PLEASE! anyone can help??????JaumeI copy the code---------------------------------------------------------------------------------------<?php$q=$_GET["q"];$con = mysql_connect('localhost', 'UsedMachines', 'jauman');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("usedmachines", $con);$sql="SELECT * FROM usedmachines WHERE family = '".$q."'";$result = mysql_query($sql);echo "<table aling=center><tr><th>Family</th><th>Model</th><th>Hours</th><th>Price</th></tr>";while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['Family'] . "</td>"; echo "<td>" . $row['Model'] . "</td>"; echo "<td>" . $row['Hours'] . "</td>"; echo "<td>" . $row['Price'] . "</td>"; echo "</tr>"; }echo "</table>";mysql_close($con);?>-------------------------------------------------------------------------------------

[HungryMind]

Hi all!!!I've adapted the code of http://www.w3schools.com/php/php_ajax_livesearch.aspto my MySQL table so I can show a live search with AJAX. Evething looks fine but when I choose the category a get an error mesage:Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\UsedMachines\getuser.php on line 25PLEASE! anyone can help??????JaumeI copy the code---------------------------------------------------------------------------------------<?php$q=$_GET["q"];$con = mysql_connect('localhost', 'UsedMachines', 'jauman');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("usedmachines", $con);$sql="SELECT * FROM usedmachines WHERE family = '".$q."'";$result = mysql_query($sql);echo "<table aling=center><tr><th>Family</th><th>Model</th><th>Hours</th><th>Price</th></tr>";while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['Family'] . "</td>"; echo "<td>" . $row['Model'] . "</td>"; echo "<td>" . $row['Hours'] . "</td>"; echo "<td>" . $row['Price'] . "</td>"; echo "</tr>"; }echo "</table>";mysql_close($con);?>-------------------------------------------------------------------------------------

IF your Database Field(Family) Nature Is Set To (Int), Then Try Without Single Quotes(')Yours: $sql="SELECT * FROM usedmachines WHERE family = '".$q."'";Mine: $sql="SELECT * FROM usedmachines WHERE family = ".$q;But This Code Is Working At My Side

[ShadowMage]

The error message means that your query failed. Try adding this code after you run the query:

$result = mysql_query($sql); //This is where you run the query//Add this codeif (!$result) {   die(mysql_error());}

That should print out the last mysql error that occured, which should tell you why the query failed.