将变量从Javascript传递到PHP并返回
问题描述：
The console ends up writing back
Uncaught ReferenceError: provhtml is not defined
Also, when I change up my Javascript to be the code right below, the console ends up just writing " <?= $provhtml ?>"
var provnum = "<?= $provhtml ?>";
console.log(provnum);
I'm not exactly sure what the problem is. I want $provhtml to be written inside the console when a user clicks a button named otherlistbut
(I'm also not sure if there needs to be the d.preventDefault();
Here is my prov.php
<?php
$phonenum = $_POST["phonenum"];
$provhtml = file_get_contents('http://api.data24-7.com/v/2.0?user=USERNAME&pass=PASSWORD&api=C&p1=1' . $phonenum);
?>
Here are my scripts in my index.html
$('button[name="otherlistbut"]').click(function(d) {
lookupProvider(d);
});
});
/************ FUNCTIONS ******************/
function lookupProvider(d) {
d.preventDefault();
var phonenum = $('input[name="phonenum"]').val();
$.ajax({
type: 'POST',
data: {
phonenum: phonenum
},
url: 'prov.php',
success: function(data) {
var provnum = $(provhtml);
console.log(provnum);
},
error: function(xhr, err) {
console.log("readyState: " + xhr.readyState + "
status: " + xhr.status);
console.log("responseText: " + xhr.responseText);
}
});
}
答
In your prov.php file
<?php
$phonenum = $_POST["phonenum"];
$provhtml = file_get_contents('http://api.data24-7.com/v/2.0?user=USERNAME&pass=PASSWORD&api=C&p1=1' . $phonenum);
echo json_encode( array( "phonenum" => $phonenum,"provhtml"=>$provhtml)); //You should be returning the values to your ajax function
?>
In your index.html
function lookupProvider(d) {
d.preventDefault();
var phonenum = $('input[name="phonenum"]').val();
$.ajax({
type: 'POST',
data: {
phonenum: phonenum
},
url: 'prov.php',
success: function(data) { //Receives the data from the php code
var provnum = data.phonenum;
console.log(provnum);
},
error: function(xhr, err) {
console.log("readyState: " + xhr.readyState + "
status: " + xhr.status);
console.log("responseText: " + xhr.responseText);
}
});
}