是否可以在 php for 循环中使用 LAST_INSERT_ID()?

Is it possible to use LAST_INSERT_ID() in php for-loop?

可以在 php for 循环中使用 LAST_INSERT_ID()。我需要在 guardian table 中获取最后一个 playground PK 作为 FK。两者需要同时插入。请原谅我没有使用 PDO,我只想让这个东西先工作。

$query = "INSERT INTO playground (parent, children, amount) VALUES ('John','susy','2000');";

$levelarray = array ("One", "Two", "Three");
for ($i = 0; $i < count($levelarray); $i++) {
    $level = $levelarray[$i];
    $query .= "INSERT INTO guardian (playgroundid, level) VALUES (LAST_INSERT_ID(),'$level');";
}
mysqli_multi_query($con, $query);

这个我也试过了。但是下面这个输出的是最后一个id,而不是新插入的id。

$query = "INSERT INTO playground (parent, children, amount) VALUES ('John','Susy','2000');";

$sql = "SELECT playground_id AS playgroundid FROM playground ORDER BY playground_id DESC LIMIT 1";
$result = mysqli_query($con, $sql);
$row = mysqli_fetch_array($result);
$playId = $row['playgroundid'];

$levelarray = array ("One", "Two", "Three");
for ($i = 0; $i < count($levelarray); $i++) {
    $level = $levelarray[$i];
    $query .= "INSERT INTO guardian (playgroundid, level) VALUES ('$playId','$level');";
}
mysqli_multi_query($con, $query);

不,这行不通。第一个守护者会得到playground的id,后面两个会得到前一个守护者的id。

改为使用 mysqli_insert_id() 获取 ID 并将其保存为 for 循环之外的变量。

您需要使用 mysqli_insert_id()

获取最后插入的 ID

然后将其用于下一个查询,如下所示:-

$query = "INSERT INTO playground (parent, children, amount) VALUES ('John','susy','2000');"; 

if(mysqli_query($con,$query)){ 
    $id = mysqli_insert_id($con);  //get last inserted id
    $levelarray = array ("One", "Two", "Three"); 
    for ($i = 0; $i < count($levelarray); $i++) { 
        $level = $levelarray[$i]; 
        $query = "INSERT INTO guardian (playgroundid, level) VALUES ($id,'$level');"; 
        mysqli_query($con,$query) ;
    } 
}

或者您仍然可以像下面这样使用 mysqli_multi_query():-

$query = "INSERT INTO playground (parent, children, amount) VALUES ('John','susy','2000');"; 

if(mysqli_query($con,$query)){ 
    $id = mysqli_insert_id($con);  //get last inserted id
    $query1 = '';
    $levelarray = array ("One", "Two", "Three"); 
    for ($i = 0; $i < count($levelarray); $i++) { 
        $level = $levelarray[$i]; 
        $query1 .= "INSERT INTO guardian (playgroundid, level) VALUES ($id,'$level');"; 
    } 
     mysqli_multi_query($con, $query1);
}

注意:- 您的查询对 SQL INJECTION 开放。所以尝试使用 prepared statement 来防止它。