我在MySql中有两个表，分别是表1和表2。因为我想通过userID将表1链接到表2。但是，我推出的功能不起作用。
MySQl 表如下:
在我的例子中，userId 将是链接这两个表的外键。 但是，我的功能不起作用。 这是我的功能如下:
function insert() {
function adduserdetails($con, $accountId, $name, $contact) {
$query = "insert into userdetails(accountId,name,contact)
values('$accountId','$name','$contact')";
//echo "{$sqlString}";
$insertResult = mysqli_query($con, $query);
if ($insertResult) {
echo " Applicant Detail Added !<br />";
echo "<a href='index.php'>Back to Home</a>";
} else {
echo " Error !";
echo "{$query}";
//header('Location: post.php');
}
}
}
if ($con->query($query) === TRUE) {
$last_id = $con->insert_id;
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $query . "<br>" . $con->error;
}
function adduseremployment($con,$last_id,$occupationMain,$comapny){
$query1 = "insert into useremployment(userId,Occupation,company)
values('$last_id',$occupationMain','$comapny')";
//echo "{$sqlString}";
$insertResult = mysqli_query($con, $query1);
if($insertResult){
echo " Applicant Detail Added !<br />";
echo "<a href='index.php'>Back to Home</a>";
}
else {
echo " Error !";
echo "{$query1}";
//header('Location: post.php');
}
}
最佳答案
您有垂直表格。
检查下面的代码，
Be remember to prepare statement.
<?php
function addUser(){
// parent table.
$queryParent = ''; // Your userdetails table insert query
$insertResult = mysqli_query($con, $queryParent);
$userId = mysqli_insert_id($con); // This is your last inserted userId.
// child table.
$queryChild = ''; // Your useremployment table insert query with userId.
$insertResult = mysqli_query($con, $queryChild);
}
?>
关于php - 在MySql php中将数据存储到带有外键的2个表中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44982940/