assigning php return value to a javascript variable

Question

I have a database form on a MySql table on which I have a javascript function to populate the options of a select tag. Those options are fetched from a table of clients who have a status of either "Active" or "Inactive", and the return values are of those clients where their status is active. In the event an order is loaded where the client status is inactive, I'm trying to add a handler for inactive clients. The form loads from a php script that left joins the client table to the main table, where clientId in the main table is equal to Id in the client table. So, I have the name and id of the client fetched outside of the function to populate the options list, regardless of their status.

There is one line that is causing me fits. I have searched this site and others and have found many solutions, but none have worked for me so far. This is the line:

var txt = <?php echo $row[`clients`.'name']; ?> ;

What is returned in Chrome and Firefox debuggers is, "Uncaught syntax error: Unexpected token <". The debugger shows: var txt = <br />

I've tried enclosing the php script in single quotes, double quotes, and without quotes, and still no luck. Any thoughts, anyone?

About an hour later--> I found a workaround. I tried all of your suggestions, but none worked in this instance. var_dump and json_encode confirmed what I knew already, that the returned data was valid. Regardless of any of the variations in syntax, they all returned the same error. What I did was to apply the same syntax as above, but in a hidden input:

<input type="text" id="cName" hidden value="<?php echo $row[`clients`.'name']?>" />

Then changed the javascript code to this:

 var txt = document.getElementById('cName').value;

Everything else works perfectly. Of course, I still have lingering thoughts about the use of backticks, and would prefer that I had a better, and safer code. As I mentioned somewhere, I simply copied the sql syntax directly from phpMyAdmin. In this instance, if I substitute single quotes for the backticks, the input returns nothing. Well, thanks all. If anyone wants to contribute more, I'll be glad to hear about it.

• Bram: Why use back ticks? Try $row['clients']['name']
• Logan Murphy: the . character means concat the strings in php...if back ticks generate a string that is.

Answer 1

That's illegal PHP syntax, and very dangerous syntax in general. Try doing a var_dump($row) to see exactly what's in that array. Probably you want something more like

var txt = <?php echo json_encode($row['clients.name']); ?>;

instead.

Note the use of json_encode(). This will ENSURE that whatever you're spitting out in the JS code block is actually syntactically valid javascript.

e.g. consider what'd happen if you ended up with

var txt = Miles O'Brien;
 ^^^^^--undefined variable;
 ^--- another undefined var
 ^--- start of a string
 ^^^^^^^---unterminated string.

with json_encode(), you end up with

var txt = "Miles O'Brien";

and everything's a-ok.

• eggyal: It's not illegal syntax... it's just almost certainly not the intended syntax!
• Marc B: you're right. forgot that backticks execute. but I highly doubt op's got a "clients" executable.
• jazcam: Guys you have me thinking and wondering. I use the same syntax on an input and it works, even if bad, illegal, or whatever. I just copied it from phpMyAdmin. I want something that works and is good code.

Answer 2

var txt = "<?php echo $row['clients']['name']; ?>";

Answer 3

var txt = <?php echo $row[`clients`.'name']; ?> ;

Consider how PHP parses this:

1. var txt = is to be output directly to the client.

2. Enter PHP mode.

3. echo the following expression.

4. Evaluate $row[`clients`.'name'].

   • First we need to determine the array index, which is the concatenation of `clients` and 'name'.

   • Backtick in PHP is the execution operator, identical to shell_exec(). So PHP attempts to execute the shell command clients, which probably fails because that isn't what you intended and it doesn't exist. Consequently, at this stage, PHP outputs an HTML error message, starting with a line break <br />.

   • Your client now has var txt = <br /> (you can verify this by inspecting the HTML source of the page returned to your browser), which it attempts to evaluate in its JavaScript context. This gives rise to the "unexpected token" error that you have witnessed.

As others have mentioned, you probably meant to do something like $row['clients']['name'] or $row['clients.name'] instead—but without seeing the rest of your PHP, it's impossible to be sure. Also, as @MarcB has observed, you need to be certain that the resulting output is valid JavaScript and may wish to use a function like json_encode() to suitably escape the value.

• jazcam: eggyal, you gave the best explanation, but still didn't get any of the solutions here to work in this instance. I did find a workaround--results below.

Answer 4

The error comes from the fact that your return value (a string in javascript) must be in quotes.

Single quotes will take whatever is between them literally, escapes (like \n ) will not be interpreted, with double quotes they will.

var txt = "<?php echo $row['clients']['name']; ?>";

is what you want

• eggyal: How can you be certain that the return value is a string and not some JavaScript expression?
• Freud Chicken: Indeed. If the developer wants to include this as well, the way to go is to jsonencode on the server, no matter what the data is. But i have the impression thats not what is happening in this particular case: $row['clients']['name'] looks pretty much like a peice of text for display (hence var txt ....)

Answer 5

Change this

var txt = <?php echo $row['clients'.'name']; ?> ;

to this:

var txt = <?php echo $row['clients']['name']; ?> ;