Question

我必须生成没有任何数字序列的Json对象，我们将如何做？我必须将此Json信息插入表中，这将在此问题中使用please find here < / p>

我的json格式

{
    "0": {
        "0": "65",
        "1": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "2": "JSS Layout, Mysore, Karnataka, India",
        "3": "Bangalore",
        "4": "ds",
        "5": "you@gmail.com",
        "6": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "7": "13-Feb-2015",
        "8": "10:30 AM",
        "9": "2",
        "10": "1220000000",
        "11": "Gajendra",
        "cId": "65",
        "address1": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "address2": "JSS Layout, Mysore, Karnataka, India",
        "city": "Bangalore",
        "comments": "ds",
        "email": "you@gmail.com",
        "landMark": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "scheduledDate": "13-Feb-2015",
        "scheduledTime": "10:30 AM",
        "services": "2",
        "userContactNumber": "1220000000",
        "userName": "Gajendra"
    }
}

我已从上方删除数字序列并更改如下

{
    "0": {
        "cId": "65",
        "address1": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "address2": "JSS Layout, Mysore, Karnataka, India",
        "city": "Bangalore",
        "comments": "ds",
        "email": "you@gmail.com",
        "landMark": "PWD Road, B Narayanapura, Bengaluru, Karnataka, India",
        "scheduledDate": "13-Feb-2015",
        "scheduledTime": "10:30 AM",
        "services": "2",
        "userContactNumber": "1220000000",
        "userName": "Gajendra"
    }
}

Myphp代码

<?php
header('Access-Control-Allow-Origin: *');//Should work in Cross Domaim ajax Calling request
mysql_connect("localhost","root","1212");
mysql_select_db("service");

$query="Select * from customer where services='2'";
$result=mysql_query($query);

if ( $result === false ) {
  die("Can\'t do that: " . mysql_error());
}

$retVal = array();
//MYSQL_ASSOC remove key =field identifier
while( $row = mysql_fetch_array( $result, MYSQL_ASSOC ) ) {
  $retVal[] = $row;
}
echo json_encode( $retVal );

Answer 1

默认情况下mysql_fetch_array()返回一个包含所有字段两次的数组

a）key =字段标识符
b）key =字段位置

告诉myqsl_fetch_array将值只在其标识符名称下放入数组中：

while( $row = mysql_fetch_array( $result, MYSQL_ASSOC ) ) {

如何生成没有任何数字序列的json对象？

1 个答案: