[英]PHP code to get MySQL row from URL param
有人可以幫我嗎?
我在某些WordPress頁面中一直在使用下面的代碼,但是很久以前我都看過它,所以我真的不記得如何調試它了-看看吧...唯一改變的是數據庫。
它是這樣的:
數據庫布局: 。 +-------+------------+------------+------------+------------+---------------+ | id | Naam | Metgesel | Kind1 | Kind2 | Email | +-------+------------+------------+------------+------------+---------------+ | abc12 | Bobby | Caily | * | * | b@example.com | | ... | ... | ... | ... | ... | ... | +-------+------------+------------+------------+------------+---------------+
發生錯誤:
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!
碼:
<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>';
}
您將需要更改以下行:
$query = sprintf("SELECT * FROM `DATABASE`.`TABLE`
WHERE idquack='$part'",
mysql_real_escape_string($query));
到這行:
$query = sprintf( "SELECT * FROM DATABASE.TABLE WHERE idquack='%s'", mysql_real_escape_string( $part ) );
您的錯誤是由於您將整個查詢本身傳遞給mysql_real_escape_string函數,而sprintf()正在尋找一個變量的事實……這一行實際上並沒有任何意義,但是我提到的是調用它的正確方法。
我沒有仔細檢查代碼的其余部分,以查看是否還有其他問題,但是請首先嘗試這樣做,以消除您給出的錯誤。
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