why does my ‘if if (isset($_POST[‘submit’])) not interact with mysql’
Please tell me, where do I go wrong, I’ve been over this a thousand times and checked the internet… It’s a lot of code, but I think I have overseen something. I made the remarks for myself, so I would’nt forget something stupid as an “;”.
Thanks in advance
http://appsolute.vacau.com/cms
<html>
<head>
<title>Milan CMS</title>
</head>
<body>
<?php
mysql_connect("mysql15.000webhost.com", "a96443****tjiran", "ev****89") or die(mysql_error()); //opening database
mysql_select_db("a9644326_app") or die(mysql_error());
$result = mysql_query("select * from milan") or die(mysql_error()); //get msql database into $result
$content = mysql_fetch_array ($result); // make $content represent current entry
?>
<form action="<?php echo $PHP_SELF;?>" method="post" name="trainingen">
<h3>Trainingen:</h3><p><textarea name="nametrainingen" rows="10" cols="60"><?php echo $content['value']; ?></textarea></p>
<input type="submit" value="Verzenden"><input type="reset" value="Veld leeg maken">
</form>
<?php
if (isset($_POST['submit']))
{ //test if submit button is clicked
mysql_query("DELETE FROM milan WHERE tag='trainingen'") or die(mysql_error()); //delete old entry
$input = $_POST['nametrainingen']; //set $input as current entry
mysql_query("INSERT INTO milan (tag, value) VALUES ('trainingen', '$input')") or die(mysql_error()); //set tag, value as trainingen and $input(current entry)
$result = mysql_query("select * from milan") or die(mysql_error()); //reset $result to new database
$content = mysql_fetch_array ($result); //make $content represent new entry
$myFile = "trainingen.json";
$fh = fopen($myFile, 'w') or die("can't open file");
fwrite($fh, json_encode($content));
fclose($fh);
}
?>
<p>
</body>
</html>
$_POSTis a super-global array that contains an associative array of all the form elements that were submitted with the form. You don’t have an input withname="submit", so yourisset()fails.