为什么我的JSON数组没有以正确的格式返回？

Question

当我回应$json2我JSON Array的回归时:

[{"contact_phonenumber":"12345"}][{"contact_phonenumber":"67890"}][{"contact_phonenumber":"123456"}][{"contact_phonenumber":"78901"}][{"contact_phonenumber":"234567"}][{"contact_phonenumber":"8901234"}]

它看起来不像JSON Array我,我确定我的循环有问题 - 也许是因为while它嵌套在for each或者什么东西,但即使我改变它,我仍然得到相同的结果.你能帮我吗？

我想要它JSON array像:

[{"contact_phonenumber":"12345"}, {"contact_phonenumber":"67890"},
 {"contact_phonenumber":"123456"}, {"contact_phonenumber":"78901"},{"contact_phonenumber":"234567"}, {"contact_phonenumber":"8901234"}]

这是我的代码:

<?php

require('dbConnect.php');

//this is the username in the user table
$Number = "+353872934480";

// get the username of the user in the user table, then get the matching user_id in the user table
// so we can check contacts against it 
$query = "SELECT * FROM user WHERE username = ?";
$stmt = $con->prepare($query) or die(mysqli_error($con));
$stmt->bind_param('s', $Number) or die("MySQLi-stmt binding failed " . $stmt->error);
$stmt->execute() or die("MySQLi-stmt execute failed " . $stmt->error);
$result = $stmt->get_result();

while($row = $result->fetch_assoc()) {

    //this is the corresponding user_id in the user table of the user
    $user_id = $row["user_id"];
}

//post all contacts for user_id as a JSON array
$phonenumberofcontact = '["+11111","+222222","12345","67890","123456","78901","234567","8901234"]';
$array = json_decode($phonenumberofcontact);

//We want to check if contacts of user_id are also users of the app. 
$query = "SELECT * FROM user WHERE username = ?";
$stmt2 = $con->prepare($query) or die(mysqli_error($con));
$stmt2->bind_param('s', $phonenumberofcontact) or die("MySQLi-stmt binding failed " . $stmt->error);

//for each value, call it $phonenumberofcontact
foreach($array as $value) {
    $phonenumberofcontact = $value;

    $stmt2->execute() or die("MySQLi-stmt execute failed " . $stmt2->error);

    //match the phone numbers in the JSON Array against those in the user table
    $result2 = $stmt2->get_result();

    while($row = $result2->fetch_assoc()) {

        //make an array called $results
        //this is a matching number in user table and in the JSON Array
        //call this username contact_phonenumber
        $results = array();

        $results[] = array(
            'contact_phonenumber' => $row['username']
        );


        $json2 = json_encode($results);
        echo $json2;

    }
}

编辑:非常感谢您的帮助,但现在当我做大部分答案建议时 - 声明$results并回显json外部while循环,我得到以下内容:

[][][{"contact_phonenumber":"12345"}][{"contact_phonenumber":"67890"}][{"contact_phonenumber":"123456"}][{"contact_phonenumber":"78901"}][{"contact_phonenumber":"234567"}][{"contact_phonenumber":"8901234"}]

你知道如何在没有空括号的情况下得到匹配的数字吗？而且,在开头和结尾只应该是方括号 - 就像JSON数组一样.

这是我更新的代码:

<?php

require('dbConnect.php');

//this is the username in the user table
$Number = "+353872934480";

// get the username of the user in the user table, then get the matching user_id in the user table
                // so we can check contacts against it 
                $query = "SELECT * FROM user WHERE username = ?";
                $stmt = $con->prepare($query) or die(mysqli_error($con));
                $stmt->bind_param('s', $Number) or die ("MySQLi-stmt binding failed ".$stmt->error);
                $stmt->execute() or die ("MySQLi-stmt execute failed ".$stmt->error);
                $result = $stmt->get_result();

            while ($row = $result->fetch_assoc()) {

            //this is the corresponding user_id in the user table of the user
            $user_id = $row["user_id"];
            }

//post all contacts for user_id as a JSON array
$phonenumberofcontact ='["+11111","+222222","12345","67890","123456","78901","234567","8901234"]';
$array = json_decode($phonenumberofcontact);

//We want to check if contacts of user_id are also users of the app. 
 $query = "SELECT * FROM user WHERE username = ?";
 $stmt2 = $con->prepare($query) or die(mysqli_error($con));
 $stmt2->bind_param('s', $phonenumberofcontact) or die ("MySQLi-stmt binding failed ".$stmt->error);

 //for each value, call it $phonenumberofcontact
    foreach ($array as $value)
    {
                $phonenumberofcontact = $value;

$stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);

//match the phone numbers in the JSON Array against those in the user table
     $result2 = $stmt2->get_result(); 
            //make an array called $results
            //this is a matching number in user table and in the JSON Array
            //call this username contact_phonenumber
$results = array();
while ($row = $result2->fetch_assoc()) {
    $results[] = array('contact_phonenumber' => $row['username']);//remove extra ,
}
$json2 = json_encode($results);
echo $json2;

    }

        ?>

Answer 1

像这样results在while循环之外声明数组:

$results = array();
foreach ($array as $value)
{
    $phonenumberofcontact = $value;

    $stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);

    $result2 = $stmt2->get_result();

    while ($row = $result2->fetch_assoc()) {
        if(!empty($row['username'])) {
            $results[] = array('contact_phonenumber' => $row['username']);//remove extra ,
        }
    }
}

$json2 = json_encode($results);
echo $json2;