[SOLVED] Extracting Database arrays

ueon · May 12, 2008

Hello

when I execute this script, the loop is incorrect.

<?php

include 'connect.php';

$css_num = mysql_query("SELECT css_file FROM stageone WHERE id = '1'");
$js_num = mysql_query("SELECT js_file FROM stageone WHERE id = '1'");
$div_num = mysql_query("SELECT div FROM stageone WHERE id = '1'");

echo "$css_num/$js_num/$div_num";

$css_int = 0;
$js_int = 0;
$div_int = 0;
$css_id = 0;
$js_id = 0;
$div_id = 0;

echo '<form name="stagetwo1" method="post" action="stagethree.php">';

while ($css_int != $css_num)
{
echo "
<p>\n
   		<input name='css_name$css_id' type='text' id='css_name$css_id' value='CSS file'>\n
</p>\n\n
";

$css_int++;
$css_id++;
}

while ($js_int != $js_num)
{
echo ("
<p>\n
   		<input name='js_name$js_id' type='text' id='js_name$js_id' value='JS file'>\n
</p>\n\n
");

$js_int++;
$js_id++;
}

while ($div_int != $div_num)
{
echo ("
<p>\n
   		<input name='div_name$div_id' type='text' id='div_name$div_id' value='DIV Image'>\n
</p>\n\n
");

$div_int++;
$div_id++;
}

echo ("
<p>
  <input type='submit' name='Submit' value='Submit'>\n
</p>\n
</form>\n
")
?>

fenway · May 13, 2008

That's a PHP issue -- I assume the mysql queries return the desired results.

Barand · May 13, 2008

Hello

when I execute this script, the loop is incorrect.

your query processing is incorrect.

replace your 3 queries with

<?php

$sql = "SELECT css_file, js_file, div 
	FROM stageone 
	WHERE id = '1'";
$result = mysql_query ($sql) or die (mysql_error());
list ($css_num, $js_num, $div_num) = mysql_fetch_row($result);      // get the data from the returned results

?>

ueon · May 14, 2008

Thank you

does that set the data into the variables?

Barand · May 14, 2008

Yes, that will put whatever was in css_file, js_file, div and put into $css_num, $js_num, $div_num

ueon · May 17, 2008

I've used the code previously provided, but I'm still having troubles :-\

"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 'div FROM stageone WHERE id = '1'' at line 1"

This is the code I'm using

<?php

include 'connect.php';

// Extracting Data
$sql = "SELECT 'css_file', 'js_file', 'div' 
	FROM stageone 
	WHERE id = '1'";
$result = mysql_query ($sql) or die (mysql_error());
list ($css_num, $js_num, $div_num) = mysql_fetch_row($result); 

//Set initial loop vars
$css_int = 0;
$js_int = 0;
$div_int = 0;
$css_id = 0;
$js_id = 0;
$div_id = 0;

//form
echo '<form name="stagetwo1" method="post" action="stagethree.php">';

//CSS loop
while ($css_int != $css_num)
{
echo "
<p>\n
   		<input name='css_name$css_id' type='text' id='css_name$css_id' value='CSS file'>\n
</p>\n\n
";

$css_int++;
$css_id++;
}

//JS Loop
while ($js_int != $js_num)
{
echo ("
<p>\n
   		<input name='js_name$js_id' type='text' id='js_name$js_id' value='JS file'>\n
</p>\n\n
");

$js_int++;
$js_id++;
}

//Div Loop
while ($div_int != $div_num)
{
echo ("
<p>\n
   		<input name='div_name$div_id' type='text' id='div_name$div_id' value='DIV Image'><BR />\n
	<input name='div_top$div_id' type='text' id='div_top$div_id' value='Top PX'><BR />\n
	<input name='div_left$div_id' type='text' id='div_left$div_id' value='Left PX'><BR />\n
</p>\n\n
");

$div_int++;
$div_id++;
}

echo ("
<p>
  <input type='submit' name='Submit' value='Submit'>\n
</p>\n
</form>\n
")
?>

ueon · May 17, 2008

The previous problem has be fixed, but now the original problem is back. the data extracted from the database is incorrect.

data in Columns

css_file = 1

js_file = 2

div = 3

table = 4

it returns:

"css_file js_file div"

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Webgen 2.0</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<div align="center">
<b>Webgen 2.0</b><br />
[stage 2]<br />
<?php

include 'connect.php';

// Extracting Data
$sql = "SELECT 'css_file', 'js_file', 'div', 'table'
	FROM stageone
	WHERE id = '1'";
$result = mysql_query ($sql) or die (mysql_error());
list ($css_num, $js_num, $div_num, $table_num) = mysql_fetch_row($result); 

echo $css_num;
echo $js_num;
echo $div_num;

//Set initial loop vars
$css_int = 0;
$js_int = 0;
$div_int = 0;
$table_int = 0;
$css_id = 0;
$js_id = 0;
$div_id = 0;
$table_id = 0;

//form
echo '<form name="stagetwo1" method="post" action="stagethree.php">';

//CSS loop
while ($css_int != $css_num)
{
echo "
<p>\n
   		<input name='css_name$css_id' type='text' id='css_name$css_id' value='CSS file'>\n
</p>\n\n
";

$css_int++;
$css_id++;
}

//JS Loop
while ($js_int != $js_num)
{
echo ("
<p>\n
   		<input name='js_name$js_id' type='text' id='js_name$js_id' value='JS file'>\n
</p>\n\n
");

$js_int++;
$js_id++;
}

//Div Loop
while ($div_int != $div_num)
{
echo ("
<p>\n
   		<input name='div_name$div_id' type='text' id='div_name$div_id' value='DIV Image'><BR />\n
	<input name='div_top$div_id' type='text' id='div_top$div_id' value='Top PX'><BR />\n
	<input name='div_left$div_id' type='text' id='div_left$div_id' value='Left PX'><BR />\n
</p>\n\n
");

$div_int++;
$div_id++;
}

//Div Loop
while ($div_int != $div_num)
{
echo ("
<p>\n
   		<input name='table_name$div_id' type='text' id='table_name$div_id' value='Table'><BR />\n
	<input name='table_width$div_id' type='text' id='table_width$div_id' value='Width'><BR />\n
	<input name='table_length$div_id' type='text' id='table_length$div_id' value='Length'><BR />\n
	<input name='table_top$div_id' type='text' id='table_top$div_id' value='Top PX'><BR />\n
	<input name='table_left$div_id' type='text' id='table_left$div_id' value='Left PX'><BR />\n
</p>\n\n
");

$div_int++;
$div_id++;
}

echo ("
<p>
  <input type='submit' name='Submit' value='Submit'>\n
</p>\n
</form>\n
")
?>
</div>
</body>
</html>

Barand · May 17, 2008

Your query selects literal string values and not column values. Remove the quotes.

ueon · May 17, 2008

I tried doing that, but it returns his error

"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 'div FROM stageone WHERE id = '1'' at line 1"

Barand · May 17, 2008

DIV is a MySQL reserved word. Put it in backticks ie `div`

or, better, change column name

ueon · May 18, 2008

thank you, it's working great now

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[SOLVED] Extracting Database arrays

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