如何同時執行多個MySql查詢？

Question

我想更新我的數據，我使用它但它不會更新我的數據， id仍然為空。 那么，如何同時執行多個 SQL 查詢呢？

$mysqli = new mysqli("a", "a", "a", "a");

if (mysqli_connect_errno()) {
   printf("Connect failed: %s\n", mysqli_connect_error());
   exit();
}

$sql = "UPDATE laporan_gini SET id_provinsi='1' WHERE nama_item_vertical_variabel= 'INDONESIA'";
$sql = "UPDATE laporan_gini SET id_provinsi='61' WHERE nama_item_vertical_variabel= 'KALIMANTAN BARAT'";

if (!$mysqli->multi_query($sql)) {
    echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

do {
    if ($res = $mysqli->store_result()) {
        var_dump($res->fetch_all(MYSQLI_ASSOC));
        $res->free();
    }
} while ($mysqli->more_results() && $mysqli->next_result());

Answer 1

你永遠不應該使用mysqli::multi_query()來執行你的 SQL，因為這個函數不支持參數綁定。 您必須使用准備好的語句分別執行這兩個查詢。 如果您不打算將任何數據綁定到您的 SQL，那么您可以將其減少一行並使用mysqli::query()代替。

如果兩個查詢相互依賴，那么您可以將它們包裝在一個事務中（假設您的表引擎支持事務）。

<?php

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("a", "a", "a", "a");
$mysqli->set_charset('utf8mb4'); // always set the charset

// start transaction
$mysqli->begin_transaction();

$sql = "UPDATE laporan_gini SET id_provinsi='1' WHERE nama_item_vertical_variabel= 'INDONESIA'";
$stmt = $mysqli->prepare($sql);
$stmt->execute();

$sql = "UPDATE laporan_gini SET id_provinsi='61' WHERE nama_item_vertical_variabel= 'KALIMANTAN BARAT'";
$stmt = $mysqli->prepare($sql);
$stmt->execute();

// commit data and end transaction
$mysqli->commit();

Answer 2

您可以在一次調用中發送多個 SQL 查詢，以分號分隔

$sql = "UPDATE laporan_gini SET id_provinsi='1' WHERE nama_item_vertical_variabel= 'INDONESIA';";
$sql.= "UPDATE laporan_gini SET id_provinsi='61' WHERE nama_item_vertical_variabel= 'KALIMANTAN BARAT';";

if (!$mysqli->multi_query($sql)) {
    echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

Answer 3

最好的方法是使用 sql 觸發器或存儲過程：
https://dev.mysql.com/doc/refman/8.0/en/trigger-syntax.html
https://dev.mysql.com/doc/connector-net/en/connector-net-tutorials-stored-procedures.html

以其他方式，您可以在代碼中運行它：

<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

if (!$mysqli->query("DROP TABLE IF EXISTS test") || !$mysqli->query("CREATE TABLE test(id INT)")) {
    echo "Table creation failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

$sql = "SELECT COUNT(*) AS _num FROM test; ";
$sql.= "INSERT INTO test(id) VALUES (1); ";
$sql.= "SELECT COUNT(*) AS _num FROM test; ";

if (!$mysqli->multi_query($sql)) {
    echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

do {
    if ($res = $mysqli->store_result()) {
        var_dump($res->fetch_all(MYSQLI_ASSOC));
        $res->free();
    }
} while ($mysqli->more_results() && $mysqli->next_result());
?>