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Hai
i have display errors ON, i looked in the php.ini and also (i have xampp) the is a phpinfo page - also ON.
But i tried the script once with a wrong db name and no error appears !?!
Perhaps its display_startup_errors (i dont know what it is) its turned off !
I changed the code like this:
Check your PHP error logs.. from what I can tell it is a parse error. Also, double check when viewing the page that NO source code is displaying.[/list]
NOTHING
when i insert: echo 'Rows returned: '. mysql_num_rows($result);
still a blank page.
and its not like my php does not work.
other skips are working!!!
Set Search Time - A google chrome extension. When you search only results from the past year (or set time period) are displayed. Helps tremendously when using new technologies to avoid outdated results.
<?php
error_reporting(E_ALL);
$db = mysql_connect("localhost","root","") or die ("MySQL-Fehler: " . mysql_error());
mysql_select_db("safe",$db) or die ("MySQL-Fehler: " . mysql_error());
$sql = "SELECT * FROM sam_artikel" or die ("MySQL-Fehler: " . mysql_error());
$result=mysql_query($sql) or die(mysql_error());
$zeile=mysql_fetch_array($result);
echo "{$zeile['eqdes']} - {$zeile['designer']} - {$zeile['value']} - {$zeile['beschreibung']} - {$zeile['eqdesignation']}";
mysql_close();
?>
But now i would like it to print out all the datasets nut just the first one!
can i do that with: $rows=mysql_num_rows($query); and than i print all rows out with WHILE... ???