JSON.parse（）不起作用 [英] JSON.parse() isn't working
问题描述
我不明白为什么我的 json.parse（）不起作用。错误可能是在我眼前，但我找不到它。以下是我的代码：
<！DOCTYPE html>
< html>
< head>
< meta charset =UTF-8>
< title> Stage galerij< / title>
< link rel =stylesheethref =css / main.css>
< script src =https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js>< / script>
< / head>
< body>
< div id =id01>< / div>
< script>
var xmlhttp = new XMLHttpRequest（）;
var url =http://www.clouddesk.be/galerij/includes/galerij_array.php;
xmlhttp.onreadystatechange = function（）{
if（xmlhttp.readyState == 4&& xmlhttp.status == 200）{
myFunction（xmlhttp.responseText）;
}
};
xmlhttp.open（GET，url，true）;
xmlhttp.send（）;
函数myFunction（响应）{
var arr = JSON.parse（response）;
var i;
var out =< table>; （i = 0; i
+
arr [i] .id +
< / td>< td> +
arr [i] .File_path +
< / td>< td> +
arr [i] .uitleg +
< / td>< / tr>;
}
out + =< / table>;
document.getElementById（id01）。innerHTML = out;
}
< / script>
< / body>
< / html>
这里是php文件的代码：
<？php
header（Access-Control-Allow-Origin：*）;
header（Content-Type：application / json; charset = UTF-8）;
$ conn = new mysqli（localhost，clouddesk_robbe，password，clouddesk_stage）;
$ result = $ conn->查询（SELECT * FROM Fotos）;
$ outp =[;
while（$ rs = $ result-> fetch_array（MYSQLI_ASSOC））{
if（$ outp！=[）{$ outp。=，;}
$ outp 。='{id：'。$ rs [idFotos]。'，';
$ outp。='File_path：'。$ rs [Foto_File]。'，';
$ outp。='uitleg：'。$ rs [Uitleg]。'，';
$ outp。='alt：'。$ rs [Alternatief]。'}';
}
$ outp。=];
$ conn-> close（）;
echo（$ outp）;
？>
它输出这个json字符串：
[{id：2，File_path：img\gallerij\Afbeelding-61.png，uitleg：Zebra verliest zijn strepen。， alt：Stressige zebra}，{id：3，File_path：img\gallerij\download（1）.jpg，uitleg：Tomatensoep meet balletjes：Nog een keer blazen ，alt：Grappige cartoon}，{id：4，File_path：img\gallerij\download.jpg，uitleg：Brilsmurf会见een raar gezicht，alt：Brilsmurf}，{id：5，File_path：img\gallerij\download（2）.jpg，uitleg：Mooi avondlijk woestijnuitzicht。，alt：Woestijnuitzicht}，{id：8，File_path：img\gallerij\image1.jpg，uitleg：Een mooi blaadje dat door het herfstseizoen van de boom is gevallen en eenzaam op de grond licht。Het is getrokken in de late herfstzon。，alt：Herfstblaadje op de grond}，{id：9，File_path：img\\ \\\\\\\\\\\\， leurd。，alt：Een kleurig paneel}，{id：10，File_path：img\gallerij\images（1）.jpg，uitleg：Mooie foto van een高尔夫死于avond getrokken。 Jeziet de reflectie van de zo in de buiging van de golf。，alt：de avondzon中的高尔夫球}，{id：12，File_path：img\gallerij\images。 jpg，uitleg：Mooie auto，het is een dure rode lexus getrokken langs de linker voorkant。，alt：Mooie rode lexus}，{id：13，File_path img\gallerij\Nikon-D810-Image-Sample-6.jpg，uitleg：Een uitzicht op een onbewoond eiland zoals je je voorstelt in je verbeelding，alt：Een onbewoond eiland}， {id：14，File_path：img\gallerij\image-slider-2.jpg，uitleg：Het huis van de film in de lucht doormiddel van duizenden ballonen aan de schoorsteen 。，alt：Huis in de lucht}，{id：15，File_path：img\gallerij\demo-image.jpg，uitleg：Een mooie foto van een zwaan。，alt：foto van een zwaan}]
一切从w3schools.com的这个页面：
http://www.w3schools.com/ JSON / json_example .asp
当我使用w3schools.com的原始代码对它进行测试时，它完美运行。但是，当我更改URL变量和数组名称时，它不再起作用。我还测试了json输出是否与在线验证器一起生效，并且它表示完全正常。
JSON.parse ）不工作，因为json字符串不正确。 File_path 是
img\gallerij\Afbeelding-61.png 但它应该是 IMG / gallerij /演示image.jpg的。请在json字符串中随处更改 File_path 。 点击 here for jsfiddle example。
File_path is从生成$ outp。='File_path：'。$ rs [Foto_File]。'，'; \\ 与 / 。
好处：您使用<$ c $在客户端直接使用 src DOM元素。
I don't understand why my json.parse() isn't working. The error is probably right before my eyes but I can't find it. Here is the code that I have:
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Stage galerij</title>
<link rel="icon" href="img/website/photo.jpg" type="image/jpg" sizes="any">
<link rel="stylesheet" href="css/main.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
</head>
<body>
<div id="id01"></div>
<script>
var xmlhttp = new XMLHttpRequest();
var url = "http://www.clouddesk.be/galerij/includes/galerij_array.php";
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
myFunction(xmlhttp.responseText);
}
};
xmlhttp.open("GET", url, true);
xmlhttp.send();
function myFunction(response) {
var arr = JSON.parse(response);
var i;
var out = "<table>";
for (i = 0; i < arr.length; i++) {
out += "<tr><td>" +
arr[i].id +
"</td><td>" +
arr[i].File_path +
"</td><td>" +
arr[i].uitleg +
"</td></tr>";
}
out += "</table>";
document.getElementById("id01").innerHTML = out;
}
</script>
</body>
</html>
And here is the code of the php file:
<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
$conn = new mysqli("localhost", "clouddesk_robbe", "password", "clouddesk_stage");
$result = $conn->query("SELECT * FROM Fotos");
$outp = "[";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
if ($outp != "[") {$outp .= ",";}
$outp .= '{"id":"' . $rs["idFotos"] . '",';
$outp .= '"File_path":"' . $rs["Foto_File"] . '",';
$outp .= '"uitleg":"' . $rs["Uitleg"] . '",';
$outp .= '"alt":"'. $rs["Alternatief"] . '"}';
}
$outp .="]";
$conn->close();
echo($outp);
?>
It outputs this json string:
[{"id":"2","File_path":"img\gallerij\Afbeelding-61.png","uitleg":"Zebra verliest zijn strepen.","alt":"Stressige zebra"},{"id":"3","File_path":"img\gallerij\download(1).jpg","uitleg":"Tomatensoep met balletjes: Nog een keer blazen en ik sla op je bek.","alt":"Grappige cartoon"},{"id":"4","File_path":"img\gallerij\download.jpg","uitleg":"Brilsmurf met een raar gezicht","alt":"Brilsmurf"},{"id":"5","File_path":"img\gallerij\download(2).jpg","uitleg":"Mooi avondlijk woestijnuitzicht.","alt":"Woestijnuitzicht"},{"id":"8","File_path":"img\gallerij\image1.jpg","uitleg":"Een mooi blaadje dat door het herfstseizoen van de boom is gevallen en eenzaam op de grond licht. Het is getrokken in de late herfstzon.","alt":"Herfstblaadje op de grond"},{"id":"9","File_path":"img\gallerij\7.jpg","uitleg":"Een kleurrijk paneel dat elke mens wel terug wat opfleurd.","alt":"Een kleurig paneel"},{"id":"10","File_path":"img\gallerij\images(1).jpg","uitleg":"Mooie foto van een golf die in de late avond getrokken is. Je ziet de reflectie van de zo in de buiging van de golf.","alt":"Golf in de avondzon"},{"id":"12","File_path":"img\gallerij\images.jpg","uitleg":"Mooie auto, het is een dure rode lexus getrokken langs de linker voorkant.","alt":"Mooie rode lexus"},{"id":"13","File_path":"img\gallerij\Nikon-D810-Image-Sample-6.jpg","uitleg":"Een uitzicht op een onbewoond eiland zoals je je voorstelt in je verbeelding","alt":"Een onbewoond eiland"},{"id":"14","File_path":"img\gallerij\image-slider-2.jpg","uitleg":"Het huis van de film up in de lucht doormiddel van duizenden ballonen aan de schoorsteen.","alt":"Huis in de lucht"},{"id":"15","File_path":"img\gallerij\demo-image.jpg","uitleg":"Een mooie foto van een zwaan.","alt":"foto van een zwaan"}]
I copied practically everything from this page of w3schools.com: http://www.w3schools.com/json/json_example.asp When I tested it with the original code of w3schools.com it worked perfectly. But when I changed the URL variable and the array names, it doesn't work anymore. I also tested if the json ouput was valid with an online validator and it said it was completely fine.
JSON.parse() is not working because json string is not proper. File_path is
img\gallerij\Afbeelding-61.png but it should be img/gallerij/demo-image.jpg. Please change File_path everywhere in json string .
Click here for jsfiddle example.
File_path is getting generated from $outp .= '"File_path":"'. $rs["Foto_File"] . '",'; So replace \ with /.
Benefit: You get a path using JSON.parse() like img/gallerij/demo-image.jpg directly in client, so that this path can be used directly as src of DOM element.
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