PHP/HTML: While loop with submit-buttons

Question

I got a select box where i can choose a Project-ID and pass it to the next form with a submit button. Right now it looks like this:

This is my code:

 <form name="form2" action="formular3.php" method="post">

 <p><strong>Choose Project:</strong></p>
 <select name = "project_id">

 <?php
 while ($row = mysqli_fetch_array($result)) {
     echo "<option value =" . $row['project_id'] . "> Project_ID: " . $row['project_id'] . " - (" . $row['name'] . ")</option>";
 }
 ?>

 </select>

 <input type="submit" value="Send" />

 </form>

Now I want to have a list which has a Submit-button for every row. It should look like this:

mickmackusa · Accepted Answer · 2017-05-24 03:43:37Z

1

If you are only sending one option per form, then remove the select tag entirely.

Better Method (and notice mysqli_fetch_assoc() instead of mysqli_fecth_array():

echo "<form name=\"form2\" action=\"formular3.php\" method=\"post\">";
    while($row=mysqli_fetch_assoc($result)){
        echo "Project_ID: {$row['project_id']} - ({$row['name']}) ";
        echo "<button name=\"project_id\" value=\"{$row['project_id']}\">Send</button>";
    }
echo "</form>";

This will submit the value in the clicked button without having to write so many form blocks into the html. You will only need to adjust the actual displaying of the buttons with <br>, table cells, etc.

My previous method that will work, but is not DRY:

while($row=mysqli_fetch_array($result)){
    echo "<form name=\"form2\" action=\"formular3.php\" method=\"post\">";
        echo "<input type=\"hidden\" name=\"project_id\" value=\"{$row['project_id']}\">";
        echo "Project_ID: {$row['project_id']} - ({$row['name']}) ";
        echo "<input type=\"submit\" value=\"Send\" />";
    echo "</form>";
}

edited May 24, 2017 at 3:43

answered May 19, 2017 at 11:58

mickmackusa♦

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1 Comment

mickmackusa Over a year ago

@k.troy2012 I gave you some suboptimal advice previously. Please implement my better method in your project.

Nithin John · Accepted Answer · 2017-05-19 10:27:28Z

1

Use submit button inside the loop,

 <p><strong>Choose Project:</strong></p>
 <select name = "project_id">

 <?php
 while ($row = mysqli_fetch_array($result)) {
     echo "<option value =" . $row['project_id'] . "> <input type='submit' value='Send' />Project_ID: " . $row['project_id'] . " - (" . $row['name'] . ")</option>";
 }
 ?>

 </select>



 </form>

answered May 19, 2017 at 10:27

Nithin John

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2 Comments

k.troy2012 Over a year ago

Looks good, but no matter what submit-button i press, it always submits the first project-ID of the list .. in this case project_id = 5

Nithin John Over a year ago

Actually You are submitting the form before selecting one from the dropdown. Just make the input type of send button from submit to button, then on change of dropdown, submit the form using jquery.

shakyl mansuri · Accepted Answer · 2017-05-19 11:01:54Z

0

You need do stop occuring current submit event and give submit button to a id, use jquery to click on submit you should post a spicific form.

like on of the submit will be clicked, you will run this script.

$(document).on('click', '#submit', function () {
    $(this).closest("form").submit();
});

answered May 19, 2017 at 11:01

shakyl mansuri

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