![](/img/trans.png)
[英]How do I access variables from PHP file using Ajax and JSON? (data is not defined error)
[英]How do i put php variables into JSON with AJAX?
我正在嘗試發送回JSON對象,但數據取決於提交AJAX調用時發送的$ id變量,我如何在JSON字符串中使用此$ id變量並將其發送回JSON.parse() ?
這基本上是我想做的,我是JSON新手
我希望它如何
$id = somenumber;
echo '{"imageSrc":"assets/img/ . $id . .jpg","second":"radi"}';
發送到JSON.parse()時應該如何
echo '{"imageSrc":"assets/img/2.jpg","second":"radi"}';
但是我不能只將變量或純文本添加到JSON嗎?
當前頁面的AJAX get請求調用PHP
require_once '../includes/db.php';
require_once '../includes/functions.php';
$dbCon = dbCon();
define("SQL", "SELECT * FROM rating ORDER BY rand() LIMIT 1");
$result = $dbCon->query(SQL);
while ($row = $result->fetch_object()) {
$id = $row->id;
//HOW DO I PARSE A VARIABLE INTO THIS JSON STRING BEFORE AJAX GETS IT??
echo '{"imageSrc":"assets/img/2.jpg","second":"radi"}';
}
編輯: json_encode()是一個php函數,感謝axiac提供的信息
<?php
$arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);
echo json_encode($arr);
?>
The above example will output:
{"a":1,"b":2,"c":3,"d":4,"e":5}
while ($row = $result->fetch_object()) {
$id = $row->id;
//in this way
echo '{"imageSrc":"assets/img/'.$id.'.jpg","second":"radi"}';
}
當用單引號引起來時,PHP不會替換變量值
$var = "some text";
echo 'hi $var'; //output : hi $var
echo "hi $var"; //output : hi some text
或者您可以將數據放入數組中並將其轉換為json這樣
$arr = ["imageSrc"=>"assets/img/".$id.".jpg","second"=>"radi"];
echo json_encode($arr);
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.