[英]Get id from other table then insert into table. I want to click the button then answer current question
<?php require_once('connection.php');
session_start();
if (isset($_POST['Submit']))
{
$answ = $_POST['answ'];
$email = $_SESSION['email'];
$insert = "INSERT INTO answerList (answ,queID,email) SELECT questionList.queID FROM questionList " ;
$result = mysqli_query($con, $insert);
if (!$result) {
echo "<script>alert('ERROR !!!!! '); location='faq.php';</script>";
}
else {
echo "<script>alert('Successful Post'); location='faq.php'; </script>";
}
}
?>
我有3個表表1:用戶(電子郵件,密碼)表2:問題(questionID,questionContent,電子郵件)表3:答案(answerID,answerContent,questionID,電子郵件)
如何將 questionID 連接到表 3? 我已經登錄並添加了問題 sql。 但是如何使用已記錄的 email 添加當前問題的答案?
您可以使用<input type='hidden' name='questionId' value='<?php echo $value>'>
對於每個問題,您必須將$value
設置為 questionID。
提交答案后,您將執行以下操作:
if (isset($_POST['Submit']))
{
$answ = $_POST['answ'];
$email = $_SESSION['email'];
$questionId = $_POST['questionId'];
$insert = "INSERT INTO `answerList`(`answer`, `email`, `questionID`) VALUES ('$answ','$email','$questionId[value-3]')" ;
$result = mysqli_query($con, $insert);
if (!$result) {
echo "<script>alert('ERROR !!!!! '); location='faq.php';</script>";
}
else {
echo "<script>alert('Successful Post'); location='faq.php'; </script>";
}
}
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