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Categorieën > PHP & SQL

SQL error (Opgelost)

blablabla - 25/09/2008 21:09
Nieuw lid		misschien klinkt t dom, want ik ben nog best wel noob in php maar wat is er fout in mijn script als ik deze error krijg: 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 '' at line 1

11 antwoorden

Gesponsorde links

Joost - 25/09/2008 21:11
PHP expert		Dat betekent dat je een fout in de syntax van je mysql-query zit. Wanneer je even je code post kunnen we kijken waar de fout zit

blablabla - 25/09/2008 21:17

Nieuw lid

php code - Bekijk de code zonder highlighting - Klap code in

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Spirit Harvest Talk</title>
</head>
<?php
session_start();
include("connect.php");
if(!(isset($_SESSION['name']))){
?>
<meta http-equiv="refresh" content="0;url=/index.php"/>
<?php
}else{
if(isset($_GET['bla'])) {
$_GET['iTalk'] = $iTalk;
$sql = 'SELECT * FROM TalkTopics WHERE TopicName = '.$iTalk;
$result = mysql_query($sql)or die(mysql_error());;
$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];
}
else
{
$sql = 'SELECT * FROM TalkTopics';
$result = mysql_query($sql)or die(mysql_error());;
$count = mysql_num_rows($result);
if($count < 20) { 
$rij = mysql_fetch_assoc($result);
echo "<a href=\"/talk.php?bla=" . $rij['TopicName'] . "\" />" . $rij['TopicName'] . "</a>";
echo "<br />";
echo $rij['TopicContent'];
}
}
}

?>

<body>
</body>
</html>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Spirit Harvest Talk</title>
</head>
<?php
session_start();
include("connect.php");
if(!(isset($_SESSION['name']))){
?>
<meta http-equiv="refresh" content="0;url=/index.php"/>
<?php
}else{
if(isset($_GET['bla'])) {
$_GET['iTalk'] = $iTalk;
$sql = 'SELECT * FROM TalkTopics WHERE TopicName = '.$iTalk;
$result = mysql_query($sql)or die(mysql_error());;
$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];
}
else
{
$sql = 'SELECT * FROM TalkTopics';
$result = mysql_query($sql)or die(mysql_error());;
$count = mysql_num_rows($result);
if($count < 20) { 
$rij = mysql_fetch_assoc($result);
echo "<a href=\"/talk.php?bla=" . $rij['TopicName'] . "\" />" . $rij['TopicName'] . "</a>";
echo "<br />";
echo $rij['TopicContent'];
}
}
}
 
?>
 
<body>
</body>
</html>

dit is de code

Joost - 25/09/2008 21:18 (laatste wijziging 25/09/2008 21:19)

PHP expert

php code - Bekijk de code zonder highlighting - Klap code in

                $sql = 'SELECT * FROM TalkTopics WHERE TopicName = '.$iTalk;

$sql = 'SELECT * FROM TalkTopics WHERE TopicName = '.$iTalk;

php code - Bekijk de code zonder highlighting - Klap code in

                $sql = "SELECT * FROM TalkTopics WHERE TopicName = '".$iTalk."' ";

$sql = "SELECT * FROM TalkTopics WHERE TopicName = '".$iTalk."' ";

blablabla - 25/09/2008 21:19 (laatste wijziging 25/09/2008 21:25)

Nieuw lid

ik heb nu dit

php code - Bekijk de code zonder highlighting - Klap code in

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Spirit Harvest Talk</title>
</head>
<?php
session_start();
include("connect.php");
if(!(isset($_SESSION['name']))){
?>
<meta http-equiv="refresh" content="0;url=/index.php"/>
<?php
}else{
if(isset($_GET['iTalk'])) {
$_GET['iTalk'] = $iTalk;
$sql = "SELECT * FROM TalkTopics WHERE TopicName = '".$iTalk."' ";
$result = mysql_query($sql)or die(mysql_error());;
$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];
}
else
{
$sql = 'SELECT * FROM TalkTopics';
$result = mysql_query($sql)or die(mysql_error());;
$count = mysql_num_rows($result);
if($count < 20) { 
$rij = mysql_fetch_assoc($result);
echo "<a href=\"/talk.php?iTalk=" . $rij['TopicName'] . "\" />" . $rij['TopicName'] . "</a>";
echo "<br />";
echo $rij['TopicContent'];
}
}
}

?>

<body>
</body>
</html>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Spirit Harvest Talk</title>
</head>
<?php
session_start();
include("connect.php");
if(!(isset($_SESSION['name']))){
?>
<meta http-equiv="refresh" content="0;url=/index.php"/>
<?php
}else{
if(isset($_GET['iTalk'])) {
$_GET['iTalk'] = $iTalk;
$sql = "SELECT * FROM TalkTopics WHERE TopicName = '".$iTalk."' ";
$result = mysql_query($sql)or die(mysql_error());;
$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];
}
else
{
$sql = 'SELECT * FROM TalkTopics';
$result = mysql_query($sql)or die(mysql_error());;
$count = mysql_num_rows($result);
if($count < 20) { 
$rij = mysql_fetch_assoc($result);
echo "<a href=\"/talk.php?iTalk=" . $rij['TopicName'] . "\" />" . $rij['TopicName'] . "</a>";
echo "<br />";
echo $rij['TopicContent'];
}
}
}
 
?>
 
<body>
</body>
</html>

maar als ik ?iTalk=De naam die die uit de databse haald
dore krijg ik niks op mn scherm, wat is er fout aan mn script?

Joost - 25/09/2008 21:26
PHP expert		php code - Bekijk de code zonder highlighting - Klap code in $_GET['iTalk'] = $iTalk; $_GET['iTalk'] = $iTalk; => php code - Bekijk de code zonder highlighting - Klap code in $iTalk = $_GET['iTalk']; $iTalk = $_GET['iTalk']; Andersom

blablabla - 25/09/2008 21:27 (laatste wijziging 25/09/2008 21:31)
Nieuw lid		alas k nu mn script draai, laat ie maar telkens 1 keer $rij['TopicName'] zien en 1 keer $rij['TopicContent'], hoe kan dit?

Joost - 25/09/2008 21:28 (laatste wijziging 25/09/2008 21:28)
PHP expert		Goed opletten !

Kr4nKz1n - 26/09/2008 08:54
Onbekend		Gebruik php code - Bekijk de code zonder highlighting - Klap code in While($rij.... While($rij....

blablabla - 26/09/2008 19:35
Nieuw lid		php code - Bekijk de code zonder highlighting - Klap code in while($rij while($rij en wat dan? tja kweet t kben best wel noob in php

Giant - 26/09/2008 20:20

PHP beginner

voorbeeldje:

php code - Bekijk de code zonder highlighting - Klap code in

$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];

$rij = mysql_fetch_assoc($result);
echo $rij['TopicName'];
echo "<br /><br />";
echo $rij['TopicName'];

Jij hebt het zo

Zo zet je hem met while loop:

php code - Bekijk de code zonder highlighting - Klap code in

while($rij = mysql_fetch_assoc($result)){
	echo $rij['TopicName'];
	echo "<br /><br />";
	echo $rij['TopicName'];
}

while($rij = mysql_fetch_assoc($result)){
	echo $rij['TopicName'];
	echo "<br /><br />";
	echo $rij['TopicName'];
}

blablabla - 27/09/2008 20:15
Nieuw lid		thnx, t werkt

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