Question

我从具有不同列的表中获取数据然后将其保存在数组中。这是我的代码：

{"allplaces":
[
{
"image":"http://www.freepngimg.com/download/lion/3-2-lion-png.png","image1":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png","image2":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png","image3":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png","image4":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png","image5":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"
}
]
}

我用json格式对此进行编码，我的出局是这样的：

     {"allplaces":[
    {"image":"http://www.freepngimg.com/download/lion/3-2-lion-png.png"},
{"image1":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"},

{"image2":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"},

{"image3":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"},

{"image4":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"},

{"image5":"http://mobileappdatabase.in/city/uploads/default-image/defaultimage.png"}]}

但我需要这样：

1. either the array's index
2. either the full string contained by the array at that index

Answer 1

如果您有1行，这只会起作用。你的桌面结构非常糟糕，但是你走了。

while ($row = mysqli_fetch_array($check,MYSQLI_ASSOC)) {
    foreach($row as $key => $value){
     $data[] = array(
       $key => $value
     );
    }    
}

如何将数组元素更改为对象？

1 个答案: