Question

How do I populate several text input fields with corresponding data, executing php-mysql query script once only?
e.g.

<form name="video" action="mysql.php" method="post" enctype="multipart/form-data" target="videofrm">
                <p>
                    <label for="url">url/gallery:</label><input class="text" type="text" id="url" name="url"/>
                    <input id="favorite" class="text" type="checkbox" name="favorite" value="true"/>
                    <label for="favorite" class="noneFloat">favorite</label>
                </p>
                <p>
                    <input id="thumb" type="file" name="thumb"/>
                </p>
                <p>
                    <label for="title">title:</label><input class="text" type="text" id="title" name="title"/>
                </p>
                <p>
                    <label for="desc">description:</label><input class="text" type="text" id="desc" name="description"/>
                </p>
                <p>
                    <label for="country">country:</label><input class="text" type="text" id="country" name="country"/>
                </p>
                <p>
                    <label for="categories">categories:</label><input class="text" type="text" id="categories" name="category"/>
                </p>
                <input type="button" value="fetch" id="fetch"/>
                <input type="reset" value="clear"/>
            </form>

I need to populate input text fields with:

select url,title,desc,country,category
from video where url='something';

First I need to know how to write this kind of php script so it returns json datatype, and afterwards how to populate text input fields, using jquery, with returned data.

`many thanks

Answer 1

Your question is very broad, and looks a bit like a “please write my code” request (although it wasn’t necessarily one). Try to avoid that impression, and mention what you have already tried/researched to avoid downvoting.

Some pointers:

• Use json_encode() to get mySQL result data into JSON form on PHP side

• Send the correct content-type header to go with it: header("content-type: application/json");

• Use jQuery’s getJSON() shorthand (or .ajax() with a json dataType) to fetch the script

• Populating the input fields with the data is trivial using .val().