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Php 从数据库输出的JSON数据_Php_Json_Database_Fetch - Fatal编程技术网

Php 从数据库输出的JSON数据

Php 从数据库输出的JSON数据,php,json,database,fetch,Php,Json,Database,Fetch,嗨,我有以下JSON数据,从我的数据库中获取: [{ "0": "1", "id": "1", "1": "Hakob", "name": "Hakob", "2": "abc@email.com", "email": "abc@email.com" }, { "0": "2", "id": "2", "1": "Arsen", "name": "Arsen", "2": "zxc@email.com",

嗨,我有以下JSON数据,从我的数据库中获取:

[{
    "0": "1",
    "id": "1",
    "1": "Hakob",
    "name": "Hakob",
    "2": "abc@email.com",
    "email": "abc@email.com"
}, {
    "0": "2",
    "id": "2",
    "1": "Arsen",
    "name": "Arsen",
    "2": "zxc@email.com",
    "email": "zxc@email.com"
}]
我不想看到以下键/值
“0”:“1”、“1”:“Hakob”、“2”:abc@email.com“,
和其他行相同。 现在由谁来做?它们是什么?我如何才能移除它们

这是我的PHP脚本,用来获取这个东西

$sql = "SELECT * FROM contacts";
$result = mysqli_query($connect, $sql);
$response = array();
while ($row = mysqli_fetch_array($result)) {
    $response[] = $row;
}

print json_encode($response);

// Close connection
mysqli_close($connect);

您只需传递mysqli_fetch_数组的第二个参数即可获得所需的结果

$sql = "SELECT * FROM contacts";
$result = mysqli_query($connect, $sql);
$response = array();

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { //<-----------change this
    $response[] = $row;
}

print json_encode($response);

// Close connection
mysqli_close($connect);
$sql=“从联系人中选择*”;
$result=mysqli\u查询($connect,$sql);
$response=array();

而($row=mysqli\u fetch\u数组($result,mysqli\u ASSOC)){//或使用
mysqli\u fetch\u ASSOC()
可能重复的