Error when echo $_GET[“jsoncallback”]

问题

<?php    
$query = mysql_query("Select id, name From table");
        while($row = mysql_fetch_array($query)) {
            echo $_GET["jsoncallback"] . '(<option value='.$row['id'].'>'.$row['name'].'</option>)';
        }
?>

When I echo result, it is error, how to fix it?

回答1:

Since you didn't write exactly what you're trying to do, i'm guessing you're trying to return a list of HTML options that a JS callback function will place in your document.

try this:

<?php    
$options = '';
$query = mysql_query("Select id, name From table");
while ($row = mysql_fetch_array($query)) {
  $options .= '<option value="'.$row['id'].'">'.$row['name'].'</option>' . "\n";
}
echo $_GET["jsoncallback"] . "('" . $options . "');";
?>

This will first create all the options as a string, and only then build the callback.

回答2:

You don't say where the error is, but presumably it is in JS:

Terminate your statements with a ;
Generate valid JS. You can't just shove a bunch of HTML in a JS function call. You need some sort of JavaScript object. json_encode will generate an Object or an Array. Or you could construct a string but escaping characters with special meaning, replacing new lines, and quoting the value.
Sanity check your $_GET data to make sure it conforms to the syntax of JS function names.

Oh, I see you have tagged this as JSON-P, in that case:

Construct a single object with all the data, convert it to JSON with encode_json, then wrap the whole thing with the callback( and );. Don't call the callback multiple times in a while loop.

回答3:

Are you sure jsoncallback is set? Try with this:

<?php
$jsoncallback = isset($_GET["jsoncallback"])? $_GET["jsoncallback"] : "";    
$query = mysql_query("Select id, name From table");
        while($row = mysql_fetch_array($query)) {
            echo $jsoncallback . '(<option value='.$row['id'].'>'.$row['name'].'</option>)';
        }
?>

回答4:

Given that you have not provided what error you are getting. I will suggest try this:

<?php  
if(isset($_GET['jsoncallback'])) 
{  
    $query = mysql_query("Select id, name From table");  
    while($row = mysql_fetch_array($query)) {  
       echo $_GET["jsoncallback"] . '(<option value='.$row['id'].'>'.$row['name'].'</option>)';
     }
}

else 
  echo "Jsoncallback not set.";
?>

回答5:

$_GET doesn't usually return errors. The possible point of error is your query.

Insert

echo mysql_error();

before your while loop;

来源：https://stackoverflow.com/questions/8134415/error-when-echo-getjsoncallback

标签

php

jsonp