how we submit php variable values with string in to database using php

Question

I want to submit drop-down with dynamic value. I concatenate all variables inside the string but data not submit any one help me how is it done

if(isset($_POST['submit'])){
     $name = $_POST['name'];
     $value1 = $_POST['value1'];
     $value2 = $_POST['value2'];


        $data = "<select name=''$name''>";
        $data .="<option>Selet data</option>";
        $data .="<option value='**$value1**'> **$value1**</option>";
        $data .="<option value='**$value2**'> **$value2**</option>";
        $data .="</select>";

        mysql_query("insert into options (value) values ('".$data."')");
    }

With variable string not submit in to the database.

HTML is below

<form action="" method="POST">

        <input type="text" name="name">
        <input type="text" name="value1">
        <input type="text" name="value2">
        <input type="submit" name="submit">

</form>

show us please the full code and your html part... btw dont use mysql_ functions they are deprecated since php 5.5.0 — Blueblazer172
– Blueblazer172, Commented Dec 7, 2016 at 12:21
You want to insert the whole shebang, including HTML tags? Or just the values? — Ionut Necula
– Ionut Necula, Commented Dec 7, 2016 at 12:22
Do query the database for errors. $data WILL contain illegal characters in the context of your query. — Florian Heer
– Florian Heer, Commented Dec 7, 2016 at 12:22
"How it is done" is to use prepared statements and query parameters. What you're building here is one great big SQL syntax error. mysql_error() is undoubtedly telling you about the problem, you should check that for errors. While not necessarily a duplicate, this would be a great place for you to start: stackoverflow.com/questions/60174/… — David
– David, Commented Dec 7, 2016 at 12:22

Rupal Javiya · Accepted Answer · 2016-12-08 05:07:35Z

0

Check with below:

    if(isset($_POST['submit'])){
     $name = $_POST['name'];
     $value1 = $_POST['value1'];
     $value2 = $_POST['value2'];

        $data = "<select name='$name'>";
        $data .="<option>Selet data</option>";
        $data .="<option value='**$value1**'> **$value1**</option>";
        $data .="<option value='**$value2**'> **$value2**</option>";
        $data .="</select>";

        if(!mysql_query("insert into options (`value`) values (".json_encode($data).")")) {
          echo("Error description: " . mysql_error());
        }
}

edited Dec 8, 2016 at 5:07

answered Dec 7, 2016 at 12:25

Rupal Javiya

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