KrazyTaco Posted March 20, 2008 Share Posted March 20, 2008 Essentially I'm trying to create a search page. I have successfully built a page that will retrieve all the entrys in the table, but I can't seem to create a page that will pull only information that relates to what a user defines. I'm using MySQL version 5.0.24a So far I have 2 pages to do this. A "search.html" page which has a box called "id" where a user can put the id of the 'ransom' they added earlier to retrieve it. This then gets sent to a page called "search.php" $id = $_POST['id']; //SQL query $query = mysql_query("SELECT * FROM `ransom_records` where id=id") or die(mysql_error()); $result = mysql_query($query) or die('Query failed: ' . mysql_error()); So what I'm trying to do is take the variable 'id' and have SQL compare the value given and find it within the database, then return only that result. When I run it, I get the following mysql_error Query failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1 So basically I'm just asking how to include a variable within a SELECT FROM WHERE statement. Thanks in advance. Quote Link to comment Share on other sites More sharing options...
rhodesa Posted March 20, 2008 Share Posted March 20, 2008 You are running mysql_query twice and need a $id...change to: <?php $id = mysql_real_escape_string($_POST['id']); $query = "SELECT * FROM `ransom_records` where id='{$id}'"; $result = mysql_query($query) or die('Query failed: ' . mysql_error()); ?> Quote Link to comment Share on other sites More sharing options...
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