php - 从数据库获取数据作为 json 响应

Question

在这里，我尝试从数据库获取一些数据，并希望将其显示为 json 响应，以便用户可以获取每个字段。

这是用户如何执行查询

http://localhost/safari/index.php?getbranch=true

这应该给出表中的分支详细信息。

这是执行此操作的 PHP 代码

<?php
    if(isset($_GET['getbranch']))
    {
        $getbranch = $_GET['getbranch'];
        if($getbranch == 'true')
        {
            getbranch($getbranch);
        }
    }
    function getbranch($getbranch)
    {
        $con = mysqli_connect('127.0.0.1', 'root', '', 'safari');
        if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
            return;
        }   
        $today = date("Ymd");           

        $result = mysqli_query($con,"SELECT division, branch,branchofficename,branchofficecode,status from tbl_branchoffice");
        while ($row = @mysqli_fetch_array($result))
        {
            $result1 = json_encode($row);             
        }
        echo $result1;
    }

这有什么问题吗？

JSON 响应:

[{"0":"3","sno":"3","1":"2","division":"2","2":"2","branch":"2","3":"SAFFARI TRAVELS","branchofficename":"SAFFARI TRAVELS","4":"gfgbhghfhf","branchofficecode":"gfgbhghfhf","5":"active","status":"active"},

{"0":"4","sno":"4","1":"2","师":"2","2":"钦奈","分行":"钦奈","3":"chennai","branchofficename":"chennai","4":"br01","branchofficecode":"br01","5":"active","status":"active"} ,{"0":"5","sno":"5","1":"3","师":"3","2":"德里","分行":"德里", "3":"德里","分支机构名称":"德里","4":"br02","分支机构代码":"br02","5":"未激活","状态":"未激活"},{ "0":"6","sno":"6","1":"2","部门":"2","2":"类加罗尔","分公司":"类加罗尔","3 ":"类加罗尔","分支机构名称":"类加罗尔","4":"br03","分支机构代码":"br03","5":"事件","状态":"事件"},{"0 ":"7","sno":"7","1":"3","部门":"3","2":"浦那","分支":"浦那","3": "pune","branchofficename":"pune","4":"br04","branchofficecode":"br04","5":"notactive","status":"notactive"}]

Answer 1

更改您的 while 循环

$result1 = array();
while ($row = @mysqli_fetch_array($result))
{    
       array_push($result1 , $row);           
 }

这样，您就收集了 $result1 中的所有结果

现在你可以对其进行编码

echo  $result1 = json_encode( $result1);  

我更喜欢使用array，忽略json_encode行代码，

foreach($result1 as $resultset){
 //resultset contains one single row of table
    foreach($resultset as $column => $columnValue){
  //assuming your table column name as 'city'
        if($column == 'city' && $columnValue == 'pune' ){
         //displaying array of table which satisfies the condition
             var_dump($resultset );

        }
    }
}