将字符串插入数组时出现问题

发布于 2024-11-02 20:54:58 字数 860 浏览 0 评论 0原文

我试图将一个内爆生成的字符串插入到一个数组中，然后用于 json 实现

内爆生成的字符串看起来像这样

'id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]

我想在这段代码中使用它

$this->_JsonArr[]=array($Generated string);

来实现类似的东西，

 $this->_JsonArr[]=array('id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]);

而不是我得到类似的

 $this->_JsonArr[]=array(" 'id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]");

东西就像生成的字符串被视为一个元素作为键和值对。显然我可以因此从 mysql 获得预期的输出，任何人都可以帮助我解决这个问题

原文

i'm trying to insert an implode generated string to an array that then later be used for json implementation

the implode generated string is look like this

'id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]

i would like to used it in this code

$this->_JsonArr[]=array($Generated string);

to achieve something like this

 $this->_JsonArr[]=array('id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]);

instead i got something like this

 $this->_JsonArr[]=array(" 'id' => $this->_SqlResult[0],'UserId' => $this->_SqlResult[1],'Msg' => $this->_SqlResult[2],'MsgStamp' => $this->_SqlResult[3]");

seem like generated string is treated as one element as key and value pair.
obviously i can get expected output from mysql because of this, can anybody help me with this

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南冥有猫 2024-11-09 20:54:58

为什么你需要内爆任何东西？只需传递数组：

$this->_JsonArr[] = your-non-imploded-array-here;

我认为您想要做的完整解决方案是这样的（即您问题中的第三个代码框）：

$row = array(
  'id' => $this->_SqlResult[0],
  'UserId' => $this->_SqlResult[1],
  'Msg' => $this->_SqlResult[2],
  'MsgStamp' => $this->_SqlResult[3]
);
$this->_JsonArr[] = $row;

Why do you need to implode anything? Just pass the array:

$this->_JsonArr[] = your-non-imploded-array-here;

I think a full solution to what you want to do is something like this (i.e., the third code box in your question):

$row = array(
  'id' => $this->_SqlResult[0],
  'UserId' => $this->_SqlResult[1],
  'Msg' => $this->_SqlResult[2],
  'MsgStamp' => $this->_SqlResult[3]
);
$this->_JsonArr[] = $row;

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