将php返回值分配给javascript变量

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.

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.

eg 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.

Answer 2

Change this

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

to this:

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

Answer 3

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

Answer 4

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.

Answer 5

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