本文介紹了Mysqli DELETE QUERY 在 PHP 腳本中不起作用的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！
問題描述
我使用以下代碼從表中刪除條目我想要做的是檢查是否從表中刪除了任何值.如果刪除了一個值，腳本應該打印成功否則為假.這就是我已經實現了.請幫助
Im using the following code to remove an entry from a table what i want to do is to check if any value was deleted from the table.If one value is deleted,the script should print success else false.This is what i have achieved till now.Please help
<?PHP
$mysqli = new mysqli("SQLHOST.COM","CLIENT","PASSWORD", "DNAME", 1234);
if ($mysqli->connect_errno) {
printf("Connect failed: %s
", $mysqli->connect_error);
exit();
}
else
{
printf("cONN Sucees");
if ($result = $mysqli->query("DELETE FROM ktable WHERE code='value'")) {
printf("Select returned %d rows.
", $result->num_rows);
printf($result->num_rows);
$result->close();
}
}
?>
推薦答案
你刪掉你需要返回的是affected_rows http://www.php.net/manual/en/mysqli.affected-rows.php
what you delete what you need to return is affected_rows http://www.php.net/manual/en/mysqli.affected-rows.php
你需要更換什么
if ($result = $mysqli->query("DELETE FROM ktable WHERE code='value'")) {
printf("Select returned %d rows.
", $result->num_rows);
printf($result->num_rows);
$result->close();
}
工作代碼
$value = ""; // Set To any Value
$mysqli = new mysqli ( "SQLHOST.COM", "CLIENT", "PASSWORD", "DNAME", 1234 );
if ($mysqli->connect_errno) {
printf ( "Connect failed: %s
", $mysqli->connect_error );
exit ();
} else {
printf ( "cONN Sucees" );
if ($mysqli->query (sprintf ( "DELETE FROM ktable WHERE code='%s'", mysqli_real_escape_string ( $mysqli, $value ) ) )) {
printf ( "Affected Rows %d rows.
", $mysqli->affected_rows );
}
}
你應該有一個工作輸出
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