MYSQL SELECT with user selected OPTION values

Question

I am trying to filter out a MYSQL table according to user selected values in my php file list.php

My table users have field stdYr, in which elements have values of S00, S01, ... S40.

in html body, I have this chunk of code:

<form method="post" role="form" action="list.php">
    <div>
        <select multiple>
            <option value="all">Select</option>
            <option value="S00">74</option>
            <option value="S01">75</option>
            <option value="S02">76</option>
            ...
            <option value="S40">114/option>
        </select>
        <button type="submit"></button>

    </div>
</form>

And then, I want to select those with selected stdYr values. The approach I could come up with is passing the values of option through $_POST:

if(isset($_POST['S00']) || isset($_POST['S11']) || ...
    ... || isset($_POST['S40'])) {
    /*some code to get only those set values*/
    $query="SELECT * FROM users WHERE stdYR = '/*some stuff*/'";
}

But with different possible selections of values, I couldn't figure out how exactly to do this... And even with the solution, I see this approach very repetitive and silly. How am I supposed to deal with such things?

I greatly appreciate your help! Thanks :D

you need to assign a name="" to <select multiple> at first. your if statement doesn't make sense. — Alex
– Alex, Commented Dec 16, 2014 at 15:41
I listed multiple Sxx to show that there's many selectable options. — Megool
– Megool, Commented Dec 17, 2014 at 12:36
@YunCHan, I've added an answer for you. Isn't it helpful for you? — A l w a y s S u n n y
– A l w a y s S u n n y, Commented Sep 16, 2018 at 12:07

Marc B · Accepted Answer · 2014-12-16 15:43:20Z

4

Your form is incorrect. Your select MUST have a name attribute for it to get submitted to the server.

<select name="options[]" multiple>
        ^^^^^^^^^^^^^^^
   <option ...>
   ...
</select>

Note the [] on the name field. That'll tell PHP to accept multiple values, making $_POST['options'] an array itself. Then it's a simple matter of:

foreach($_POST['options'] as $option) {
     insert_into_db($option);
}

answered Dec 16, 2014 at 15:43

Marc B

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A l w a y s S u n n y · Accepted Answer · 2014-12-16 15:44:10Z

3

Use this way:

<form method="post" role="form" action="list.php">
 <div>
    <select name="MyOptions[]" multiple>
        <option value="all">Select</option>
        <option value="S00">74</option>
        <option value="S01">75</option>
        <option value="S02">76</option>
        ...
        <option value="S40">114/option>
    </select>
    <button type="submit"></button>

  </div>
</form>

 print '<pre>';
 print_r($_POST);
 $selectOption = $_POST['MyOptions']; //get post result and then work as you wish

answered Dec 16, 2014 at 15:44

A l w a y s S u n n y

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Comments

Alex · Accepted Answer · 2014-12-16 16:06:44Z

-1

You could try this :

<select name="stdYR[]" multiple>

<?php
$stdYR = $_POST['stdYR'];
$stdYR_ct = count($stdYR);
$i=0;
while($i < $stdYR_ct){
    $stdYR_SQL .= "'$stdYR[$i]',";
    $i++;
}
if (preg_grep("/All/i", $stdYR) or ($stdYR_ct < 1)){
    $stdYR_SQL = "";
}else{
    $stdYR_SQL = eregi_replace(",$",'',$stdYR_SQL);
    $stdYR_SQL = " AND stdYR IN($stdYR_SQL)";
}

$query = "SELECT * FROM users WHERE 1=1 $stdYR_SQL ";
?>

edited Dec 16, 2014 at 16:06

answered Dec 16, 2014 at 15:55

Alex

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1 Comment

tadman Over a year ago

No, you don't just throw arbitrary strings in your query, especially not if they're from $_POST.

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