Question

Assuming this that we in time insert data in database has 3 input:radio and with select one from they, insert value it to database. now how in select from database same radio that inserted in database, is checked.

Example:
We have 3 input:radio as:

<input type="radio" name="type1" value="value1">
<input type="radio" name="type2" value="value2" checked>
<input type="radio" name="type3" value="value3">

With checked value2 inserted it in database, now we want show(select) all radios and checked it radio that is inserted to database.as(this is after select from database):

<input type="radio" name="type1" value="value1">
<input type="radio" name="type2" value="value2" checked> // this value was in the database
<input type="radio" name="type3" value="value3">

How can fix it with PHP?

Answer 1

You should, as alreay said, use the same name for the radio buttons. Also, the right sintax is checked="checked" :

<input type="radio" name="type" value="value1">Value 1
<input type="radio" name="type" value="value2" checked="checked">Value 2
<input type="radio" name="type" value="value3">Value 3

When retrieved from database, you can check if value equals to the one in your html, and set a checked attribute accordingly.

<input type="radio" name="type" value="value1" <?php echo ($query_hs->type == 'value1') ? 'checked="checked"' :'';?>>Value 1
<input type="radio" name="type" value="value2" <?php echo ($query_hs->type == 'value2') ? 'checked="checked"' :'';?>>Value 2
<input type="radio" name="type" value="value3" <?php echo ($query_hs->type == 'value3') ? 'checked="checked"' :'';?>>Value 3

Just substitute your own values with my standard one, like:

<input type="radio" name="type" value="hotel" <?php echo ($query_hs->type == 'hotel') ? 'checked="checked"' :'';?>>Hotel