从数据库获取数据作为json响应

Question

在这里，我试图从数据库中获取一些数据，并希望将其显示为json响应，以便用户可以获取每个字段。

这是用户执行查询的方式

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

这应该提供表中的分支详细信息。

这是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;
    }

这怎么了？

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”，“ division”：“ 2”，“ 2”：“ chennai”，“ branch”：“ chennai”，“ 3“：” chennai“，” branchofficename“：” chennai“，” 4“：” br01“，” branchofficecode“：” br01“，” 5“：” active“，” status“：” active“}，{” 0“：” 5“，” sno“：” 5“，” 1“：” 3“，” division“：” 3“，” 2“：” delhi“，” branch“：” delhi“，” 3“ ：“ delhi”，“ branchofficename”：“ delhi”，“ 4”：“ br02”，“ branchofficecode”：“ br02”，“ 5”：“ notactive”，“ status”：“ notactive”}，{“ 0” ：“ 6”，“ sno”：“ 6”，“ 1”：“ 2”，“除法”：“ 2”，“ 2”：“班加罗尔”，“分支”：“班加罗尔”，“ 3”：“ bangalore“，” branchofficename“：” bangalore“，” 4“：” br03“，” branchofficecode“：” br03“，” 5“：” active“，” status“：” active“}，{” 0“：” 7“，” sno“：” 7“，” 1“：” 3“，”除“：” 3“，” 2“：”浦那“，”分支“：”浦那“，” 3“：”浦那“ ，“ branchofficename”：“浦那”，“ 4”：“ br04”，“ branchofficecode”：“ br04”，“ 5”：“无效”，“状态”：“无效”}]

Answer 1

$result1 = array();
if(isset($_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[] = $row;             
    }
    echo json_encode($result1);
}

首先将结果的每一行收集到数组result1中，然后最后输出$ result1数组的json_encode

Answer 2

更改while循环

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

这样做，您已经在$result1收集了所有结果

现在您可以对其进行编码

echo  $result1 = json_encode( $result1);  

我将更喜欢使用array ，ignor 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 );

        }
    }
}