I did also creating a different class. but still not working...
but if I place the '<script>console.log("message here");</script>' will work..
//index.html
assuming that this is a plete code.
<form action="post.php" method="post">
<input type="text" name="name" id="name"/>
<input type="submit" value="submit"/>
</form>
//post.php
<?PHP
if(isset($_POST['name'])){
echo "<script>console.log('".$_POST['name']."');</script>";
}
?>
my problem is , i cant use console.log in submitting a form. but if I did this in redirection It will work..
The Function under my class is ...
public function console($data) {
if(is_array($data) || is_object($data)) {
echo("<script>console.log('".json_encode($data)."');</script>");
} else {
echo("<script>console.log('".$data."');</script>");
}
}
I did also creating a different class. but still not working...
but if I place the '<script>console.log("message here");</script>' will work..
//index.html
assuming that this is a plete code.
<form action="post.php" method="post">
<input type="text" name="name" id="name"/>
<input type="submit" value="submit"/>
</form>
//post.php
<?PHP
if(isset($_POST['name'])){
echo "<script>console.log('".$_POST['name']."');</script>";
}
?>
my problem is , i cant use console.log in submitting a form. but if I did this in redirection It will work..
The Function under my class is ...
public function console($data) {
if(is_array($data) || is_object($data)) {
echo("<script>console.log('".json_encode($data)."');</script>");
} else {
echo("<script>console.log('".$data."');</script>");
}
}
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edited Jan 7, 2016 at 3:15
Donnie Gallocanta Jr.
asked Jan 7, 2016 at 1:37
Donnie Gallocanta Jr.Donnie Gallocanta Jr.
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12
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So is the problem you're having that your
console.log
isn't firing when you load it into the browser with PHP? – Nick Zuber Commented Jan 7, 2016 at 1:40 -
Perhaps input the
console.log
argument as String?echo '<script>console.log("'.$_POST['name'].'");</script>';
– E. Sundin Commented Jan 7, 2016 at 1:41 - not actually load the php.. but when I submitting a form. – Donnie Gallocanta Jr. Commented Jan 7, 2016 at 1:41
- <script>console.log(".$_POST['name'].");</script> is working when I place in a index.html but my problem is when I place this into the target of a form – Donnie Gallocanta Jr. Commented Jan 7, 2016 at 1:43
-
@DonnieGallocantaJr. If you place the code directly in HTML it should work. I imagine it would not work within PHP because of the missing
"
around the String argument. In my previous ment I've included the necessary"
. – E. Sundin Commented Jan 7, 2016 at 1:48
3 Answers
Reset to default 3It does not work within PHP because of the missing " around the String argument of console.log
.
The output would've been
<script>console.log(name);</script>
instead of
<script>console.log("name");</script>
Solution
echo '<script>console.log("'.$_POST['name'].'");</script>';
If you are trying to debug or see the value that was posted from the front end to back end then you can simply use the chrome inspector.
- Right click anywhere in browser and click
inspect element
. - click on
network
tab. - submit your form with desired values.
- on the left click on the
post.php
. - Click on the
headers
on the right and scroll down to findForm Data
. - You will have all your post variables listed there with respective values.
You seem to be trying to debug the $_POST
variable, If thats the case, then Please note, that console.log()
is a frontend debugging tool used in javascript
, and has nothing to do with php
.
Few good way of checking the content of variables in php
.
1. print_r
This will output the content of a variable that can be an array
, object
or string
in a nice human readable format.
echo '<pre>';
print_r($_POST);
echo '</pre>';
die();
2. var_dump
This will output the content of the variable with extra respective details like datatypes
and length
.
var_dump($_POST);
die();