我有两张 table
<jobs> { jobID(p-k) , JobName }
<Account> {SSU(p-k),emploeeName}
和 连接表在上面的表之间建立关系,
<JobLists> { No(p-k) , jobID(f-k) , SSU(f-k) }
并且我想在用户单击“pick”时将数据插入到
if( isset($_POST['pick']) ){
if( empty($_POST['JobId']) || $_POST['JobId'] == 0 ){
echo"<h4> choose something </h4>";
}else{
echo "what do u want to do..!! ";
include('../CIEcon.php'); //$dbCIE
$impid = implode("' , '" , $_POST['JobId']);
echo $impid;
$sqlInsert ="INSERT INTO JobsLists(`JobID` , `SSU`) VALUES(".$impid.",'$SSU' )";
$MyQuery= mysqli_query($dbCIE, $sqlInsert) or die(mysqli_error($dbCIE));
// TEST ONLY ////////----------------------------------------////////////
if (mysqli_affected_rows($dbCIE) > 0) {
echo "You have successfully added a job.<br><br>";
}else{"Error occurred when trying to add a job. <br> " ; }
////////----------------------------------------////////////
最佳答案
你不能只循环作业 ID 吗?
foreach($_POST['JobID'] as $impid) {
$sqlInsert = "INSERT INTO JobLists(`JobID` , `SSU`) VALUES(".$impid.",'$SSU' )";
}
关于php - 将多条记录插入连接表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35120736/