scaleautostyle Posted July 2, 2010 Share Posted July 2, 2010 Hi guy's!! I try several option but always get the same result.. doesn't work and receive this message Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\...........\SEARCH.php on line 15 this is my form <?php include_once("db_connection.php"); ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <title>Search engine</title> </head> <p><body> <h3>Model kit database</h3> <p>You may search either by reel car or kit manufacturer</p> <form method="post" action="search.php?go" id="searchform"> <input type="text" name="name"> <input type="submit" name="submit" value="Search"> </form> </body> </html> </h2> <p> </p> <p> </p> <br> <br> </body> </html> and this is my search code <?php include_once("db_connection.php"); if(isset($_POST['submit'])){ if(isset($_GET['go'])){ if(preg_match("/^[ a-zA-Z]+/", $_POST['name'])){ $name=$_POST['name']; //-query the database table $sql="SELECT kit_id, description, manufacturer_kit FROM Contacts WHERE description LIKE '%" . $name . "%' OR manufacturer_kit LIKE '%" . $name ."%'"; //-run the query against the mysql query function $result=mysql_query($sql); //-create while loop and loop through result set while($row=mysql_fetch_array($result)){ $description =$row['description']; $manufacturer_kit=$row['manufacturer_kit']; $kit_id=$row['kit_id']; //-display the result of the array echo "<ul>\n"; echo "<li>" . "<a href=\"search.php?id=$kit_id\">" .$description . " " . $manufacturer_kit . "</a></li>\n"; echo "</ul>"; } } else{ echo "<p>Please enter a search query</p>"; } } } ?> what is wrong??? thanks for your help sebastien Link to comment https://forums.phpfreaks.com/topic/206559-why-its-not-working/ Share on other sites More sharing options...
PFMaBiSmAd Posted July 2, 2010 Share Posted July 2, 2010 The error means that your query failed due to an error (could be a connection problem, a database problem, a table problem, a column problem, or a syntax problem.) If you search for any part of that error message, you find that using mysql_error(); will tell you why the query failed. For troubleshooting purposes (remove it after you are done) echo mysql_error(); on the next line after the line with your mysql_query(...); statement. Link to comment https://forums.phpfreaks.com/topic/206559-why-its-not-working/#findComment-1080450 Share on other sites More sharing options...
scaleautostyle Posted July 2, 2010 Author Share Posted July 2, 2010 thanks but now I have an other problem. when I submit the search form I got a BLANK page.... ?????? what's going on here ??? the form page does'nt change just the search.php page. ( VERY MINOR ) and I got the same result if I put back the old one. ???? <?php include_once("db_connection.php"); if(isset($_POST['submit'])){ if(isset($_GET['go'])){ if(preg_match("/^[ a-zA-Z]+/", $_POST['name'])){ $name=$_POST['name']; //-query the database table $sql="SELECT kit_id, description, manufacturer_kit FROM Contacts WHERE description LIKE '%" . $name . "%' OR manufacturer_kit LIKE '%" . $name ."%'"; //-run the query against the mysql query function $result=mysql_query($sql); //-create while loop and loop through result set while($row=mysql_fetch_array($result)){ $description =$row['description']; $manufacturer_kit=$row['manufacturer_kit']; $kit_id=$row['kit_id']; //-display the result of the array echo "<ul>\n"; echo "<li>" . "<a href=\"search.php?id=$kit_id\">" .$description . " " . $manufacturer_kit . "</a></li>\n"; echo "</ul>"; } } else{ echo "<p>Please enter a search query</p>"; } } } ?> did my coding is good ??? Link to comment https://forums.phpfreaks.com/topic/206559-why-its-not-working/#findComment-1080482 Share on other sites More sharing options...
scaleautostyle Posted July 3, 2010 Author Share Posted July 3, 2010 OK PROBLEM SOLVE... now a new one... see other post. Link to comment https://forums.phpfreaks.com/topic/206559-why-its-not-working/#findComment-1080676 Share on other sites More sharing options...
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