First want to start off by saying that I am still a beginner developer but have gotten a long way in a short time and I am somewhat stumped. And yes I know my code might not be pretty in layout, but still learning, at least things are working.
I am creating something that is like a client portal for shows. A client signs up to do a show from an intake form. When they submit the form, it goes to Monday.com, creates a folder and sub folder in dropbox and then inserts everything into my Mysql database. I also then have another page (Assets) where they can upload files based on the show. Now if they have signed up for multiple shows, at the top of this page I have a dropdown box that grabs all the shows that is assigned to their used id. When they click on the show that they want to add files to and then click the "Choose Show For Asset Upload" button it goes back to the database to retrieve the dropbox path and the file request url and puts it into the code where those variables are assigned. So, everything is working great except when it comes to a show that has a single quote (apostrophe). I noticed this when I added a test show and everything went bonkers. I was able to figure everything out when it comes to making it correct in the code for Monday and Dropbox and even INSERTing it into the database. In the database column it has "Michael's" instead of "Michael\'s", so it's exactly how it should be in there. In the dropdown it actually shows "Michael's" but yet when I do an echo after clicking the button it shows "Michael" So that single quote is definitely the issue and this is where I don't know how to fix it, after much searching through the net.
In the dropdown it lists (Show Test, Did I Really Do It!!, Dudley, The Amazing MA, Lets See If This Works, Michael's).
Code is:
<div class="topdiv">
<h2 style="text-align: center;">Assets Upload</h2><br>
</div>
<?php
$userid = $_SESSION["userid"];
$sql = "SELECT * FROM intake WHERE userid = ?;";
$stmt = mysqli_stmt_init($con);
if(!mysqli_stmt_prepare($stmt, $sql)){
echo "There was an internal error!";
exit();
} else {
mysqli_stmt_bind_param($stmt, "s", $userid);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
}
?>
<form class="formcenter" action="assets.php" method="post">
<select name="show" id="show">
<?php
while ($rows = mysqli_fetch_assoc($result)) {
$show = $rows['showtitle'];
echo "<option value='$show'>$show</option>";
}
?>
</select>
<input type="submit" name="chooseShow" value="Choose Show For Asset Upload"><br><br>
</form>
<?php
if(isset($_POST["chooseShow"])) {
$showTitle = $_POST["show"]; //WHEN I DO AN ECHO OF THIS IT SHOWS "MICHAEL" NOT "MICHAEL'S"
$sql = "SELECT * FROM intake WHERE userid = ? and showtitle = ? ;";
$stmt = mysqli_stmt_init($con);
if(!mysqli_stmt_prepare($stmt, $sql)){
echo "There was an internal error!";
exit();
} else {
mysqli_stmt_bind_param($stmt, "ss", $userid, $showTitle);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$rows = mysqli_fetch_assoc($result);
$path = $rows['dbxPath'];
$requestURL = $rows['requestURL'];
echo $show; //THERE JUST TO SEE WHAT IT WAS OUTPUTTING
echo $showTitle; //THERE JUST TO SEE WHAT WAS OUTPUTTING
}
}
?>
When I do the two echos at the end of the code $show = "Michael's" and $showTitle = "Michael". So I do know that the correct way is coming through but just can't grab it to put in the last $sql variable to use as the $showTitle Granted, I am assuming the $show is showing "Michael's" because it's the last show in the loop. BUT when I tested $show instead of $showTitle in the mysqli_stmt_bind_param statement it actually worked, so I know it's possible. Just need to know how to get the full "Michael's" into my $showTitle variable.
Thank you for taking the time to look through this longwinded (trying to give you as much info as possible) question and appreciate any help and advice.
-Michael
