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10.06.2008 at 09:42AM PDT, ID: 23791015
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9.3

mysql_result is not working correctly when sending variable as field name.

Asked by MeridianManagement in PHP and Databases, MySQL Server, SQL Query Syntax

Tags:

I'm trying to display a bunch of fields from a table to edit.

towards the bottom of this code, I use $value[name] which is the field name. But then when I stick it inside mysql_result($result,0,$value[name]) it comes out blank. I know the $result is good because I replace $value[name] with a static field name and $value[name] is good because they both echo out. They just don't seem to work together.

For some reason, mysql_result isn't accepting $value[name]. Is it the way I've created $value? Help!Start Free Trial 
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// get field names and comments
	@mysql_select_db("information_schema",$mysql_db) or die( "Unable to select the database. (Schema)");
	$query = " SELECT COLUMN_COMMENT, COLUMN_NAME, COLUMN_TYPE
	FROM COLUMNS WHERE COLUMN_COMMENT != ''
	AND TABLE_NAME = '$primary_key_table'
	AND TABLE_SCHEMA LIKE '$client_dbname'
	ORDER BY COLUMN_COMMENT ";
	$temp = mysql_query ( $query,$mysql_db) or die ( "Sql error : " . mysql_error( ) );
	$num_fields = mysql_num_rows ( $temp );
	
	// get field names and types
	for ( $i = 0; $i < $num_fields; $i++ ) 
	{
		if(mysql_result($temp,$i,"COLUMN_COMMENT")!='') 
			$sql_fields[$i][description] .= mysql_result($temp,$i,"COLUMN_COMMENT"); // comment found
		else $sql_fields[$i][description] .= mysql_result($temp,$i,"COLUMN_NAME"); // comment not found, use fieldname
		$sql_fields[$i][name] .= mysql_result($temp,$i,"COLUMN_NAME");
		$sql_fields[$i][type] .= mysql_result($temp,$i,"COLUMN_TYPE"); // build list of types for determining edit box
	}
	
	// display data
	@mysql_select_db($client_dbname,$mysql_db) or die( "Unable to select the database. (Client)");
	$query = "SELECT * FROM $primary_key_table WHERE $primary_key = '$_REQUEST[Edit]' LIMIT 1";
	$result = mysql_query($query,$mysql_db) or die(mysql_error());
	mysql_num_rows($result);
	?>
	<form action="<?=$_SERVER['PHP_SELF']?>" method="get"><table width="100%" style="background-color:#EEEEEE; border: 1px solid #CCCCCC;">
    <table width="100%" style="background-color:#EEEEEE; border: 1px solid #CCCCCC;"><tr><td valign="top"><?
	
	$count=0;
	foreach ($sql_fields as &$value)  //iterate each field within query
	{
		$count++;
		if($count%2==1) $bgcolor="#FFFFFF";
		else 			$bgcolor="#EEEEEE";
 
		//$value = str_replace( '"' , '""' , $value );
		?><div style="border: 1px solid #CCCCCC; background: <?=$bgcolor?>; padding: 10px;">
		<?=$value[description]?><br><?
		
		// determine input type
		if(substr_count($value[type],'varchar'))
			{ 
				?><input  
				name="<?=$value[name]?>" 
				style="width: 97%; font-size:10px; background: <?=$bgcolor?>" 
				type="text" 
				value="<? 
				$temp = trim($value[name]);
				echo mysql_result($result,0,$temp); ?>"><?
				echo $value[name];			
			} 
		
		?></div><?
	}
 
	?></td></tr></table>
    </form><?
[+][-]10.06.2008 at 09:56AM PDT, ID: 22652000

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[+][-]10.06.2008 at 10:01AM PDT, ID: 22652049

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[+][-]10.06.2008 at 10:02AM PDT, ID: 22652057

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[+][-]10.06.2008 at 11:39AM PDT, ID: 22653002

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About this solution

Zones: PHP and Databases, MySQL Server, SQL Query Syntax
Tags: php mysql
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Solution Provided By: hielo
Participating Experts: 1
Solution Grade: A
 
 
[+][-]10.06.2008 at 12:03PM PDT, ID: 22653228

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[+][-]10.06.2008 at 12:19PM PDT, ID: 22653399

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