在MySQL php中将ID从一个表插入到另一个表

我有两个表格，一个是

信息

，另一个是

劳动力

信息

`

劳动力

调用

addinfo（）

时，我希望将数据插入到信息中，并且自动递增的

id

将插入表workForce，列twf

这就是我尝试过的

addinfo.php

<?php 

    if($_SERVER['REQUEST_METHOD']=='POST'){

        //Getting values
        $name = $_POST['name'];
        $weather = $_POST['weather'];
        $date = $_POST['date'];
        $status = $_POST['status'];
        $timeIn = $_POST['timeIn'];
        $timeOut = $_POST['timeOut'];

        //Creating an sql query
        $sql = "INSERT INTO information(name, weather, date, status, time_in, time_out) VALUES ('$name','$weather','$date', '$status', '$timeIn', '$timeOut')";
        $sql="INSERT INTO work_force (twf) VALUES (LAST_INSERT_ID(), )"


        //Importing our db connection script
        require_once('dbConnect.php');

        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo 'Information Added Successfully';
        }else{
            echo 'Could Not Add Information';
        }

        //Closing the database 
        mysqli_close($con);
    }
?>

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){

        //Getting values
        $subcontractors = $_POST['subcontractors'];
        $noPeople = $_POST['noPeople'];
        $noHours = $_POST['noHours'];
        $twf = $_POST['twf'];


        //Creating an sql query
        $sql = "INSERT INTO work_force(subcontractors, number_of_person, number_of_hours, twf) VALUES ('$subcontractors','$noPeople','$noHours','$twf')";

        //Importing our db connection script
        require_once('dbConnect.php');

        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo 'Work Force Added Successfully';
        }else{
            echo 'Could Not Add Work Force';
        }

        //Closing the database 
        mysqli_close($con);
    }
?>

addWorkForce.php

<?php 

    if($_SERVER['REQUEST_METHOD']=='POST'){

        //Getting values
        $name = $_POST['name'];
        $weather = $_POST['weather'];
        $date = $_POST['date'];
        $status = $_POST['status'];
        $timeIn = $_POST['timeIn'];
        $timeOut = $_POST['timeOut'];

        //Creating an sql query
        $sql = "INSERT INTO information(name, weather, date, status, time_in, time_out) VALUES ('$name','$weather','$date', '$status', '$timeIn', '$timeOut')";
        $sql="INSERT INTO work_force (twf) VALUES (LAST_INSERT_ID(), )"


        //Importing our db connection script
        require_once('dbConnect.php');

        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo 'Information Added Successfully';
        }else{
            echo 'Could Not Add Information';
        }

        //Closing the database 
        mysqli_close($con);
    }
?>

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){

        //Getting values
        $subcontractors = $_POST['subcontractors'];
        $noPeople = $_POST['noPeople'];
        $noHours = $_POST['noHours'];
        $twf = $_POST['twf'];


        //Creating an sql query
        $sql = "INSERT INTO work_force(subcontractors, number_of_person, number_of_hours, twf) VALUES ('$subcontractors','$noPeople','$noHours','$twf')";

        //Importing our db connection script
        require_once('dbConnect.php');

        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo 'Work Force Added Successfully';
        }else{
            echo 'Could Not Add Work Force';
        }

        //Closing the database 
        mysqli_close($con);
    }
?>

---

您的代码中存在一些问题，如果不执行第一个查询，所有最后插入的id都不能像这样插入

您可以这样使用：

//Creating an sql query
$sql = "INSERT INTO information(name, weather, date, status, time_in, time_out) VALUES ('$name','$weather','$date', '$status', '$timeIn', '$timeOut')";

//Importing our db connection script
require_once('dbConnect.php');

//Executing query to database
if(mysqli_query($con,$sql)){
    echo 'Information Added Successfully';
    $lastid = mysqli_insert_id();       
    $sql = "INSERT INTO work_force (subcontractors, number_of_person, number_of_hours, twf) VALUES ('$subcontractors','$noPeople','$noHours',$lastid)";
    mysqli_query($con,$sql);
}else{
echo 'Could Not Add Information';
}

您没有执行第一个sql，在哪里执行？除了第二个sql中的额外逗号，您必须确保同时运行这两个sql，而不仅仅是覆盖变量的第二个sql。@Jon再次查看了我的帖子。在此之前，我分别添加了它们，但是我无法将
id
插入“twf”…@GoudaElalfy再次查看我的帖子。在此之前，我分别添加了它们，但是我无法将id插入“twf”……这个逗号
LAST\u INSERT\u id（），
感谢您的回答。如果调用
addWorkForce（）
，数据是否会插入到插入
twf
的同一行中？或者它将插入另一行？是的，它是。。现在时间是你的了试一下别忘了我的旁注。。这对美国来说更重要@为什么你撤销了我把mysql函数改成mysqli函数的操作？这与mysqli_uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu^^