JSON保存在数据库中并使用JQuery加载

I create a huge JSON-Object and save it in my database. But when I load the "string" and echo it in PHP, I can't access the JSON Object in JQuery. Do I have to consider something if I want to save my JSON Object in a MySQL Database (when I just create the Array and then echo it with "echo json_encode($arr);" it works fine, but I need to save the Object for caching).

我创建了一个巨大的JSON-Object并将其保存在我的数据库中。但是当我加载“string”并在PHP中回显它时,我无法访问JQuery中的JSON对象。如果我想将我的JSON对象保存在MySQL数据库中(当我只是创建数组然后用“echo json_encode($ arr)”回显它时,我是否必须考虑一下;“它工作正常,但我需要保存对象用于缓存)。

{"247":{"0":"This is a question","1":"","2":"247","3":"0","answers":[["Answer1","960","1"],["Answer 2","962","0"],["Answer 3","961","0"],["Answer 4","963","0"]]},{"248":{"0":"This is a question","1":"","2":"247","3":"0","answers":[["Answer1","960","1"],["Answer 2","962","0"],["Answer 3","961","0"],["Answer 4","963","0"]]}}

{“247”:{“0”:“这是一个问题”,“1”:“”,“2”:“247”,“3”:“0”,“答案”:[[“Answer1”, “960”,“1”],[“答案2”,“962”,“0”],[“答案3”,“961”,“0”],[“答案4”,“963”,“ 0“]]},{”248“:{”0“:”这是一个问题“,”1“:”“,”2“:”247“,”3“:”0“,”答案“: [[“Answer1”,“960”,“1”],[“答案2”,“962”,“0”],[“答案3”,“961”,“0”],[“答案4” , “963”, “0”]]}}

just an excerpt

只是一个摘录

If I just echo this JSON-Object, everything works fine, but if I load the same string from the database and echo it, it doesn't work.

如果我只是回显这个JSON-Object,一切正常,但如果我从数据库加载相同的字符串并回显它,它就不起作用。

Update 1: forget to tell that I'm using a TEXT-Field with UTF8_general_ci collation

更新1:忘记告诉我我正在使用带有UTF8_general_ci排序规则的TEXT-Field

Update 2: Maybe a little bit more code:

更新2:也许更多代码:

function start() {
    $(".start").click(function () {

        $.post("load_script.php", { }, function(data){
            alert(data[247][0]);
        }, "json");

        return false;
    });
}

this loads the script and should alert "This is a question"

这会加载脚本,并应提醒“这是一个问题”

<?php
require_once('connect.php');

$ergebnis = mysql_query("SELECT text FROM cache_table ORDER BY RAND() LIMIT 1");
while($row = mysql_fetch_object($ergebnis)) {
    $output = $row->text;
}

echo $output;

this is the script, where I load the database entry with the JSON-Object.

这是脚本,我用JSON-Object加载数据库条目。

Update 3: I think I solved the problem. Some break sneaked into my JSON-Object so I do this, before the output:

更新3:我想我解决了这个问题。有些打破潜入我的JSON-Object所以我这样做,在输出之前:

$output = str_replace("\n", "", $output);
$output = str_replace("\r", "", $output);
$output = str_replace("\r\n", "", $output);

3 个解决方案

#1

I'd suggest looking at what your javascript is seeing. Instead of asking jQuery to interpret the json for you, have a look at the raw data:

我建议看看你的javascript看到了什么。不要让jQuery为你解释json,而是查看原始数据:

function start() {
    $(".start").click(function () {

        $.post("load_script.php", { }, function(data){
                alert(data);
        }, "text");

        return false;
    });
}

For example, if part of the string gets oddly encoded because of the UTF-8, this might cause it to appear.

例如,如果由于UTF-8而导致部分字符串被奇怪编码,则可能导致它出现。

Once you've done that, if you still can't spot the problem, try this code:

完成后,如果仍然无法发现问题,请尝试以下代码:

var data1, data2;
function start() {
    $(".start").click(function () {

        $.post("load_script.php", {src: "db" }, function(data){
                data1 = data;
        }, "text");

        $.post("load_script.php", {src: "echo" }, function(data){
                data2 = data;
        }, "text");

        if (data1 == data2) {
           alert("data1 == data2");
        }
        else {
           var len = data1.length < data2.length ? data1.length : data2.length;
           for(i=0; i<len; ++i) {
              if (data1.charAt(i) != data2.charAt(i)) {
                 alert("data1 first differs from data2 at character index " + i);
                 break;
              }
           }
        }

        return false;
    });
}

And then change the PHP code to either return the data from the database or simply echo it, depending on the post parameters:

然后更改PHP代码以从数据库返回数据或简单地回显它,具体取决于post参数:

<?php
   if ($_POST['src'] == 'db')) {
      require_once('connect.php');

      $ergebnis = mysql_query("SELECT text FROM cache_table ORDER BY RAND() LIMIT 1");
      while($row = mysql_fetch_object($ergebnis)) {
        $output = $row->text;
      }
   }
   else {
      $output = '{"247":{"0":"This is a question","1":"","2":"247","3":"0","answers":[["Answer1","960","1"],["Answer 2","962","0"],["Answer 3","961","0"],["Answer 4","963","0"]]},{"248":{"0":"This is a question","1":"","2":"247","3":"0","answers":[["Answer1","960","1"],["Answer 2","962","0"],["Answer 3","961","0"],["Answer 4","963","0"]]}}';
   }

echo $output;
?>

Hope that helps!

希望有所帮助!

3 个解决方案

#1

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