php

country_language

Answer 1

实际上，在更新过程中，您将获得$ id作为国家ID，并且可能在表country_language中； country_id不是NULL，这就是为什么要更新其工作状态

进入代码的插入部分或其他部分，$ id不保存任何内容，并且您也没有将新插入的国家/地区ID的值分配给$id，因此使代码不起作用

您可以做的只是通过在if条件检查$POST['language']

之前使用下面的代码来获取插入到国家/地区表中的行的最后一个插入ID。

$id = mysqli_insert_id($connect);

然后使用此$id插入country_language表中，它将起作用

Answer 2

// Using prepared statements you dont need to do this, and its safer    
// $name = mysqli_real_escape_string($connect, $_POST["name"]);  
// $description = mysqli_real_escape_string($connect, $_POST["description"]); 

$id = isset($_POST["country_id"]) ? $_POST["country_id"] : '';

if($id != '') {  
    // Update query   
    . . .

    $message = 'Data Updated';    
} else {  
    $connect->begin_transaction();
    $sql = "INSERT INTO country (name, description) 
                VALUES(?,?)");
    // prepare the query and bind paramters to the ?
    $ins = $conect->prepare($sql);
    $ins->bind_params('ss', $_POST["name"], $_POST["description"]);
    // execute the query
    $res = $ins->execute();

    // test to see if insert worked
    if ( ! $res ) {
        // insert failed.
        echo $connect->error;
        // no rollback required as nothing has been updated anyway
        exit;
    }


    // capture the new id created by above INSERT
    $new_ctry_id = $connect->insert_id;

    // Using a foreach loop so it will run 
    // for each occurance in $_POST["language"]
    // so only need to check it exists

    if(isset($_POST["language"])) {
        // prepare the statement outside the loop 
        // and execute it with new params as many times as you like
        $sql = "INSERT INTO country_language 
                    (country_id, language_id) VALUES(?,?)";
        $ins2 = $connect->prepare($sql);

        foreach($_POST["language"] as $lang) {
            // bind the new parameters each time round the loop
            $ins2->bind_params('is', $new_ctry_id, $lang);

            $res = $ins2->execute();

            // test to see if insert worked
            if ( ! $res ) {
                echo 'Insert of languages failed with ' . $connect->error;
                // run rollback to remove the insert of the country
                $connect->rollback();
                exit;
            }
        }
    }

    // if we get here, all is well and we must commit the changes we made
    $connect->commit();

    $message = 'Data Inserted';
}

如何插入PHP 2表中？

php

2 个答案: