jquery how to break string and continue string after - php

im trying to do this to get loop i variable inside my php variable
var monthyear[i] = "<?php echo $startmonth_name"+i+" ?>";
so it will show like:
var monthyear[i] = "<?php echo $startmonth_name1 ?>";
var monthyear[i] = "<?php echo $startmonth_name2 ?>";
can't figure it out and i keep getting error:
Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';'

Loop in javascript won't work for reasons mentioned in my comment, you can try having a loop in php code, something like below.
Replace this with
var monthyear[i] = "<?php echo $startmonth_name1 ?>";
var monthyear[i] = "<?php echo $startmonth_name2 ?>";
With
<?php
for($i=0;$i<some_value;$i++)
{
echo "var monthyear[".$i."] = ". ${'startmonth_name'.$i} .";"
}
?>
Note: I haven't tested the code. This is just to give an idea.

do this instead:
<?php echo "var monthyear["+i+"] = '"+ $startmonth_name[i]+"';"; ?>;
change these to an array to use them in the loop: $startmonth_name1

Forget about JavaScript right now. JavaScript will be executed later, in another computer, possibly some thousand kilometres away. You are getting a PHP parse error: that means that your PHP code does not even run. And it's clear why—the PHP block contained in this code:
var monthyear[i] = "<?php echo $startmonth_name"+i+" ?>";
... is this:
echo $startmonth_name"+i+"
You might get away with something like this:
var monthyear[i] = <?php echo json_encode($startmonth_name); ?>+i;
Yet I suggest you remember that PHP and JavaScript will simply neither interact nor share variables and rethink your logic accordingly.

Related

2 php vars with json encode in javascript

I have a problem with my script. If I have two JSON_ENCODE lines my script doesn't work anymore.
var time= <?php echo json_encode($time_cell); ?>;
var time2= <?php echo json_encode($time_cell2); ?>;
Why? Is there any other method to get this PHP vars?
JSON encoded string needs to be parsed in Javascript.
You can do somthing like this:
var time=JSON.parse("<?php echo json_encode($time_cell); ?>");
var time2=JSON.parse("<?php echo json_encode($time_cell2); ?>");
You mey read more here on this link.

Printing PHP variables into JavaScript variables [duplicate]

This question already has answers here:
Pass a PHP string to a JavaScript variable (and escape newlines) [duplicate]
(14 answers)
Closed 9 years ago.
Yes I know this question gets asked a lot, but I'm fairly new to JS and I need to use a php variable in some JS. I'm more then aware that PHP is executed server side and JS is client side however other people claim that this works.
I've got a PHP variable called "test1" that I want to log to the JS console (for instance):
<?php
$test1 = '1';
print '
<script type="text/javascript">
var carnr;
carnr = "<?php print($test1); ?>"
console.log(carnr);
</script>';
?>
What this does is print " " to the JS console. Not exactly what I was hoping for.
Now this may not even be doable and I may have to pass the variable off the page and back in again with AJAX, but I'd rather have a quick and easy solution if there is one available!
Any help is appreciated.
You could do this.
<script>
var JSvar = "<?= $phpVar ?>";
</script>
The PHP will be parsed and the value of $phpVar will become the value of JSvar whatever.
Make sure you encode phpVar properly. For example, if phpVar contains a double quote, you'll end up with a broken JS
Use this no need to give "" => change to '.$test1.'..
<?php
$test1 = '1';
print '
<script type="text/javascript">
var carnr;
carnr = "'.$test1.'"
console.log(carnr);
</script>';
?>
try
<?php $test1 = '1'; ?>
<script type="text/javascript">
var carnr;
carnr = "<?php print($test1); ?>"
console.log(carnr);
</script>
Generally, it is better to not print static stuff with php, but to have static (that is unchanging) stuff directly in HTML and only use PHP on the parts that really need it.
You made a mistake do it so:
<?php
$test1 = '1';
echo '<script type="text/javascript"> var carnr; carnr = "'.$test1.'" console.log(carnr)</script>';
?>
Since you're writing your JS with your PHP, you can simply do:
$test1 = "blah";
echo "<script type=\"text/javascript\">console.log($test1);</script>";
You are already in open php tag. When printing the line just append the variable to the output by using a dot.
Example:
print 'variable1 '.$variable1.' is now printed';

passing PHP string to JavaScript: unterminated string literal

i am assigning PHP var to my javascript var and sending to PHP file through
ajax-jQuery, but my php variable contains newline chars which
i have replaced with <br>
e.g. $values1 = 'abc<br>pqr<br>xyz'; $values2 = 'xyz<br>lmn';
javascript - var data = 'val1=<?php echo $values; ?>&val2=<?php echo $values2; ?>';
and then ajax script to post data to PHP file
but when i print this data on console it is giving me error- SyntaxError: unterminated string literal.
Can anyone help ?
Your JS code:
var data = 'val1=<?php echo $values1; ?>&val2=<?php echo $values2; ?>';
Will give Javascript syntax error if one or more of your PHP variables $values1 OR $values2 contain single quote ' in them.
Make sure your PHP variable don't contain single quotes in them by replacing all single quotes to something else otherwise use double quotes " to create JS var like this:
var data = "val1=<?php echo $values1; ?>&val2=<?php echo $values2; ?>";
Provided PHP variables don't contain double quotes.
First of all, you have a typo, that might cause an error:
// --------------------------------v
var data = 'val1=<?php echo $values1; ?>&val2=<?php echo $values2; ?>';
Then, I suggest you to use object as data parameter for Ajax request:
var data = {
val1: '<?php echo $values1; ?>',
val2: '<?php echo $values2; ?>'
};
Also it is better to escape single quotes ' in both $values1 and $values2 variables.
Try using <br /> instead of <br>. Just guessing here, no testing.

transfer a Variable from php to js

I got really confused when i tried to transfer variables from php to js.
everything is fine when i try to get value using an id
for example:
var name = $("#name").val();
but my question is, if i want to convert for example this variable:
$id = 182938; //the id of the user
to this variable:
var id;
this question might be dumb... maybe easy for you, but not for me :p
I have looked it up and only found something like this anywhere i looked for:
<?php
$number = rand();
?>
<script language="javascript">
var num = <?php echo $number ?>
</script>
Does it have the same affect as passing down the value?
Thanks!!
Not quite. Encoding as JSON will ensure that it is a valid JavaScript literal.
var num = <?php echo json_encode($number) ?>
what the code does is:
<?php
$number = rand(); //assign random number to $number. let's say it's "3"
?>
//then these are printed on the page normally
<script language="javascript">
var num = <?php echo $number ?> //the $number is echoed at this position
</script>
and then the browser receives:
<script language="javascript">
var num = 3;
</script>
You have to generate syntactically VALID javascript. PHP can echo text into ANY part of the page it's generating, but that doesn't necessarily mean it's actually USABLE text. In your case, you've forgotten a ;, so your generated javascript is actually a syntax error:
var num = <?php echo $number ?>;
^--- missing
The same sort of thing occurs when you're outputting text. Consider where you're storing someone's name into a variable:
<?php
$name = "Miles O'Brien"; // <-- note the single quote
?>
<script>
var name = <? php echo $name ?>;
</script>
This LOOKS ok, but it'll actually generate yet another syntax error:
var name = Miles O'Brien;
there's no quotes around the whole string, and the embedded quote will START a string in javascript, leading the JS interpreter to complain about an undefined variable "Miles", an undfined variable "O", and an unterminated string literal.
Instead, the PHP should have been:
var name = <? echo json_encode($name) ?>;
The JSON encapsulation guarantees that you're embedding syntactically valid Javascript values.

how to pass array string in JavaScript function from PHP end as a argument?

I am getting the error missing ) after argument list in my Firebug console.
emissing ) after argument http://a8.sphotos.ak.fbcdn.net/hphotos-ak-snc7/s720x720/393131_320846714645076_100001592501599_911297_470580896_n.jpg
My question is how to pass $char_data variable in JavaScript function as a argument
Define php variable:
<?php
$chart_data = "['NBA',1],['NFL',2],['MLB',3],['NHL',4]";
$div = "graph";
?
Call JavaScript function with define argument
<script>
dynamicChartArray('<?php echo $div;?>','<?php echo $chartdata;?>')
</script>
A function of JavaScript
<script>
function dynamicChartArray(div,chartdata){
var myData = new Array(chartdata);
var myChart = new JSChart(div, 'pie');
alert(chartdata+div);
}
<script>
Rather than creating an array out of a string in javascript, why not just just get the PHP to output it as an array to start with?
Just add an extra set of [] which javascript reads as an array.
$chart_data = "[['NBA',1],['NFL',2],['MLB',3],['NHL',4]]";
then ditch the quotes on the output (which are responsible for causing the error messages)
dynamicChartArray('<?php echo $div;?>', <?php echo $chartdata;?>);
and then myData can just equal chart data (since its already an array)
var myData = chartdata;
'<?php echo $chartdata;?>'
This is going to echo '['NBA',1],['NFL',2],['MLB',3],['NHL',4]'. Note how there are single quotes inside the single quotes.
new Array(chartdata)
This will just make an array, with one element, the string "['NBA',1],['NFL',2],['MLB',3],['NHL',4]".
Try doing dynamicChartArray('<?php echo $div;?>',[<?php echo $chartdata;?>])
This will make chartdata an array of arrays.
Instead of
$chart_data = "['NBA',1],['NFL',2],['MLB',3],['NHL',4]";
Use
$chart_data = "[\"NBA\",1],[\"NFL\",2],[\"MLB\",3],[\"NHL\",4]";
Change your call to this:
dynamicChartArray('<?php echo $div;?>',[<?php echo $chartdata;?>])
And function to this:
function dynamicChartArray(div,chartdata){
var myData = chartdata;
var myChart = new JSChart(div, 'pie');
alert(chartdata+div);
}
change this:
dynamicChartArray('<?php echo $div;?>','<?php echo $chartdata;?>')
to this:
dynamicChartArray('<?php echo $div;?>', [<?php echo $chart_data;?>]);
and see if it works
You dont need var myData = new Array(chartdata);.
chartdata is already an array.
Take a look at json_encode.
$chart_data = json_encode(array(array('NBA',1),array('NFL',2)));
which will produce a json string ready to echo into your script
string(21) "[["NBA",1],["NFL",2]]"
You should have a look at the output. I bet it is:
dynamicChartArray('graph','['NBA',1],['NFL',2],['MLB',3],['NHL',4]')
and you can already see that you have problems with the quotes.
Instead of creating a string, I suggest to create an array and use json_encode:
$chart_data = array(
array('NBA',1),
array('NFL',2),
array('MLB',3),
array('NHL',4)
);
and
dynamicChartArray('<?php echo $div;?>', <?php echo json_encode($chartdata); ?>)
JSON happens to be valid JavaScript as well and it gives you more possibilities to process the data on the server side.

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