tallberg Posted February 14, 2007 Share Posted February 14, 2007 Im attemping to pass a $condtion variable from a hidden field back to the mysql_query. I can pass number but not value with quotes. This is the code so far: <? echo '<input type="hidden" name="query" value="' . $condition . '">'; ?> Which is passed to if($condition == ''){ $condition = $_POST["query"]; echo "this is the condition ". $condition ." END "; } result from my echo statement and the sql error: this is the condition The_Usage LIKE ''%bog Garden, Pond%'' END Problem reading table here!: 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 '%bog Garden, Pond%''' at line 1 Thanks if anyone can help Quote Link to comment Share on other sites More sharing options...
ToonMariner Posted February 14, 2007 Share Posted February 14, 2007 add the quotes in the query.... Quote Link to comment Share on other sites More sharing options...
JJohnsenDK Posted February 14, 2007 Share Posted February 14, 2007 post the whole script... atleast post the mysql_query code you are using.. Quote Link to comment Share on other sites More sharing options...
tallberg Posted February 15, 2007 Author Share Posted February 15, 2007 What is happening is <? echo '<input type="hidden" name="query" value="' . $condition . '">'; ?> posts back The_Group=''Bamboo'' some how an extra set of single quotes are posted arround the property. Quote Link to comment Share on other sites More sharing options...
tallberg Posted February 15, 2007 Author Share Posted February 15, 2007 Found a solution. <input type='hidden' name='query' value='<? echo urlencode($condition); ?>'> $con = $_POST["query"]; $condition = urldecode($con); Quote Link to comment Share on other sites More sharing options...
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