PHP:无法编码多行的json [英] PHP: can't encode json with multiple rows

问题描述

在发布问题之前，我已经花了几个小时仔细研究了几个类似的答案.

I've spent a couple of hours looking through several the similar answers before posting my problem.

我正在从数据库中的表中检索数据，我想将其编码为JSON.但是，只有表具有一行时，json_encode()的输出才有效.如果有多个行，则 http://jsonlint.com/处的测试将返回错误.

I'm retrieving data from a table in my database, and I want to encode it into a JSON. However, the output of json_encode() is only valid when the table has one single row. If there is more than one row, the test at http://jsonlint.com/ returns an error.

这是我的查询:

$result = mysql_query($query);

    $rows = array();

    //retrieve and print every record
    while($r = mysql_fetch_assoc($result)){
        $rows['data'] = $r;

        //echo result as json
        echo json_encode($rows);
    }

这将为我提供以下JSON:

That gets me the following JSON:

{
"data": 
    {
        "entry_id":"2",
        "entry_type":"Information Relevant to the Subject",
        "entry":"This is my second entry."
    }
}


{
"data":{
        "entry_id":"1",
        "entry_type":"My Opinion About What Happened",
        "entry":"This is my first entry."
    }
 }

当我在 http://jsonlint.com/上运行测试时，它会返回以下错误:

When I run the test at http://jsonlint.com/, it returns this error:

    Parse error on line 29:
    ..."No comment"    }}{    "data": {    
    ---------------------^
    Expecting 'EOF', '}', ',', ']'

但是，如果我仅使用JSON的前半部分...

However, if I only use this first half of the JSON...

{
"data": 
    {
        "entry_id":"2",
        "entry_type":"Information Relevant to the Subject",
        "entry":"This is my second entry."
    }
}

...或者如果我仅测试下半部分...

... or if I only test the second half...

{
    "data":{
        "entry_id":"1",
        "entry_type":"My Opinion About What Happened",
        "entry":"This is my first entry."
    }
 }

...同一测试将返回有效JSON".

... the same test will return "Valid JSON".

我想要的是能够在表的每一行中以单个[有效] JSON输出.

What I want is to be able to output in one single [valid] JSON every row in the table.

任何建议将不胜感激.

推荐答案

问题是您要为每行吐出单独的JSON，而不是一次完成所有操作.

The problem is you're spitting out separate JSON for each row, as opposed to doing it all at once.

$result = mysql_query($query);

$rows = array();

//retrieve and print every record
while($r = mysql_fetch_assoc($result)){
    // $rows[] = $r; has the same effect, without the superfluous data attribute
    $rows[] = array('data' => $r);
}

// now all the rows have been fetched, it can be encoded
echo json_encode($rows);

我所做的较小更改是将数据库的每一行作为新值存储在$rows数组中.这意味着完成后，您的$rows数组将包含查询的所有所有行，因此一旦完成，您就可以获得正确的结果.

The minor change I've made is to store each row of the database as a new value in the $rows array. This means that when it's done, your $rows array contains all of the rows from your query, and thus you can get the correct result once it's finished.

解决方案的问题是，您正在为数据库的某一行回显有效的JSON，但是json_encode()并不知道所有其他行，因此您将获得一系列单独的JSON对象，例如与包含一个数组的单个数组相对.

The problem with your solution is that you're echoing valid JSON for one row of the database, but json_encode() doesn't know about all the other rows, so you're getting a succession of individual JSON objects, as opposed to a single one containing an array.

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