有人能帮我吗?
我一直在我的一些WordPress页面中使用下面的代码,但我很久以前就看过了,我真的不记得如何调试它了——想想吧…唯一改变的是数据库。
它是这样工作的:
- URL的参数名为id,格式如下:http://example.com/post?id=...
- 代码会检查是否存在param,否则会重定向到主页
- 如果存在param,代码将获取ID并将其与ISP托管的MySQL数据库中的记录进行比较
- Match用于echo语句
- 页面上的div被激活
数据库布局:.+-------+------------+------------+------------+------------+---------------+
| id | Naam | Metgesel | Kind1 | Kind2 | Email |
+-------+------------+------------+------------+------------+---------------+
| abc12 | Bobby | Caily | * | * | b@example.com |
| ... | ... | ... | ... | ... | ... |
+-------+------------+------------+------------+------------+---------------+
遇到错误:
Warning: mysql_real_escape_string() expects parameter 1 to be string, array given in /home/.../public_html/wp-content/plugins/insert-php-code-snippet/shortcode-handler.php(32) : eval()'d code on line 4
Invalid or no security key!
代码:
<script>
function invite(){
document.getElementById('invite').style.display=(document.getElementById('invite').style.display=='block')?'none':'block';
}
</script>
<script>
function returnHome(){
setTimeout(function () {window.location.href = 'http://example.com';},2000);
}
</script>
$part = $_REQUEST['id'];
if(isset($_GET["id"])){
$query = sprintf("SELECT * FROM `DATABASE`.`TABLE`
WHERE idquack='$part'",
mysql_real_escape_string($query));
$result = mysql_query($query);
if (!$result) {
$message = 'Invalid or no security key';
die($message);
} else {
while ($row = mysql_fetch_assoc($result)) {
if ($row['Metgesel'] != "*"){
if ($row['Metgesel'] == "#"){
if ($row['Kind1'] != "*"){
if ($row['Kind2'] != "*"){
echo '<h1>' . $row['Naam'] . ", " . "Metgesel" . ", " . $row['Kind1'] . " en " . $row['Kind2'] . "</h1>";
} else {
echo '<h1>' . $row['Naam'] . ", " . "Metgesel" . " en " . $row['Kind1'] . "</h1>";
}
} else {
echo '<h1>' . $row['Naam'] . " en " . "Metgesel" . "</h1>";
}
} else{
if ($row['Kind1'] != "*"){
if ($row['Kind2'] != "*"){
echo '<h1>' . $row['Naam'] . ", " . $row['Metgesel'] . ", " . $row['Kind1'] . " en " . $row['Kind2'] . "</h1>";
} else {
echo '<h1>' . $row['Naam'] . ", " . $row['Metgesel'] . " en " . $row['Kind1'] . "</h1>";
}
} else {
echo '<h1>' . $row['Naam'] . " en " . $row['Metgesel'] . "</h1>";
}
}
} else {
echo '<h1>' . $row['Naam'] . "</h1>";
}
echo '<script>invite();</script>';
}
}
mysql_free_result($result);
} else{
echo 'Hold on tight - we're taking you home!';
echo '<script>returnHome();</script>';
}
您需要更改此行:
$query = sprintf("SELECT * FROM `DATABASE`.`TABLE`
WHERE idquack='$part'",
mysql_real_escape_string($query));
到此行:
$query = sprintf( "SELECT * FROM DATABASE.TABLE WHERE idquack='%s'", mysql_real_escape_string( $part ) );
您的错误是由于您将整个查询本身传递给mysql_real_eescape_string函数,而sprintf()正在查找一个变量。。。这句话确实没有任何意义,但我提到的方式是调用它的正确方式。
我没有仔细查看其余的代码,看看是否还有其他问题,但请先尝试删除您给出的错误。