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PHP varible from mysql query in form, when null

I have an html form used to edit a record in a mysql table.
In the case of editing, I prefill the form with the current values for the record as returned by the query, as below:

<INPUT TYPE="text" NAME="SITENAME" SIZE=35 VALUE="<?php echo $customer['sitename']; ?>">

I would like to use the same form for creating a new record, but of course there would be no values assigned to the sql query array, and as such the form would not work.
I know I could put an IF or ISSET for every input, or a single one but then have to duplicate the form code without each echo statement. Both seem cumbersome.

Is there a better way to use the same form code?

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2

Answers


  1. 0

    I think that the most compact way is:

    value="<?php echo $customer['sitename'] ?? ''; ?>"

    Additionally, You should sanitize the value, it can contain harmful characters like ":

    value="<?php echo htmlspecialchars($customer['sitename'] ?? ''); ?>"
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  2. 0

    The null coalesce operator is perfect for this situation. It’s basically shorthand for if(isset()):

    You’re thinking of doing something like this:

    if (isset($customer['sitename'])) {
    echo $customer['sitename'];
} else {
    echo $someDefaultValue;
}

    This can be shortened to:

    echo $customer['sitename'] ?? $someDefaultValue;

    And since you want nothing in the default case, you can just use null:

    echo $customer['sitename'] ?? null;

    I.e., output the existing sitename if you have it, else nothing. And note you can use the short open syntax if all you’re doing is echo, so your code becomes:

    <INPUT TYPE="text" NAME="SITENAME" SIZE=35 VALUE="<?= $customer['sitename'] ?? null; ?>">
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