How to avoid Query doubles INSERT into database? - php

Question

Problem:
I have made a simple form that uses PHP to pass information to my database via a INSERT query. However, every time I run it, it tries to put the information in twice. How can I avoid this?
Explanation:
I first insert the answers, into my answers table, save the AnswerID as a variable. Then do the save with my question table and lastly I use the two saved variables containing the ID's into my question_answers table.
My code:
if (isset($_POST['textinput1']) && !empty($_POST['textinput1'])) {
$text1 = mysqli_real_escape_string($conn, $_POST['textinput1']);
$text2 = mysqli_real_escape_string($conn, $_POST['textinput2']);
$q_text = mysqli_real_escape_string($conn, $_POST['textarea']);
$stmt = $conn->prepare("INSERT INTO answers (Answer1Text, Answer2Text) VALUES (?, ?)");
$stmt->bind_param('ss', $text1, $text2);
$stmt->execute();
$answerid = $stmt->insert_id;
$stmt = $conn->prepare("INSERT INTO question (QuestionText) VALUES (?)");
$stmt->bind_param('s', $q_text);
$stmt->execute();
$questionid = $stmt->insert_id;
if ($stmt->execute()) {
$stmt = $conn->prepare("INSERT INTO question_answers (AnswerID, QuestionID) VALUES (?, ?)");
$stmt->bind_param('ss', $answerid, $questionid);
$stmt->execute();
echo "<h2>Dit spørgsmål er nu lagt op på siden!</h2>";
echo "<h3>Tusinde tak for din interesse for SMIL - Skodfri Århus.</h3>";
}
else
{
echo "ERROR: Could not able to execute . " . mysqli_error($conn);
}
}
// close connection
mysqli_close($conn);
?>
My tables of importance:
question: QuestionID(PK), QuestionText
answers: AnswerID(PK), Answer1Text, Answer2Text
question_answers: QuestionAnswerID(PK), QuestionID(FK), AnswerID(FK)
Ps. I prefer not to use composite unique constraint as a solution.
Also a side-question, should $stmt->insert_id variables be mysqli_real_escape_string?

Answer 1

Your problem is that you have executed the second query TWICE
if (isset($_POST['textinput1']) && !empty($_POST['textinput1'])) {
$text1 = mysqli_real_escape_string($conn, $_POST['textinput1']);
$text2 = mysqli_real_escape_string($conn, $_POST['textinput2']);
$q_text = mysqli_real_escape_string($conn, $_POST['textarea']);
$stmt = $conn->prepare("INSERT INTO answers (Answer1Text, Answer2Text) VALUES (?, ?)");
$stmt->bind_param('ss', $text1, $text2);
$stmt->execute();
$answerid = $stmt->insert_id;
$stmt = $conn->prepare("INSERT INTO question (QuestionText) VALUES (?)");
$stmt->bind_param('s', $q_text);
$stmt->execute();
$questionid = $stmt->insert_id;
// THIS IS THE SECOND EXECUTION OF QUERY 2
if ($stmt->execute()) {
$stmt = $conn->prepare("INSERT INTO question_answers (AnswerID, QuestionID) VALUES (?, ?)");
$stmt->bind_param('ss', $answerid, $questionid);
$stmt->execute();
echo "<h2>Dit spørgsmål er nu lagt op på siden!</h2>";
echo "<h3>Tusinde tak for din interesse for SMIL - Skodfri Århus.</h3>";
}
else
{
echo "ERROR: Could not able to execute . " . mysqli_error($conn);
}
}
// close connection
mysqli_close($conn);
?>
Instead try this as the IF test
//if ($stmt->execute()) {
if ( isset($answerid,$questionid) ) {

Answer 2

if ($stmt->execute()) {
this runs one of your statements a second time. You should assign the return value to a variable if you need it for something later.