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[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?
我正在尝试发送回JSON对象,但数据取决于提交AJAX调用时发送的$ id变量,我如何在JSON字符串中使用此$ id变量并将其发送回JSON.parse() ?
这基本上是我想做的,我是JSON新手
我希望它如何
$id = somenumber;
echo '{"imageSrc":"assets/img/ . $id . .jpg","second":"radi"}';
发送到JSON.parse()时应该如何
echo '{"imageSrc":"assets/img/2.jpg","second":"radi"}';
但是我不能只将变量或纯文本添加到JSON吗?
当前页面的AJAX get请求调用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"}';
}
编辑: json_encode()是一个php函数,感谢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"}';
}
当用单引号引起来时,PHP不会替换变量值
$var = "some text";
echo 'hi $var'; //output : hi $var
echo "hi $var"; //output : hi some text
或者您可以将数据放入数组中并将其转换为json这样
$arr = ["imageSrc"=>"assets/img/".$id.".jpg","second"=>"radi"];
echo json_encode($arr);
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