Question

我已经看到了一些类似的问题，并尝试了修复，但它们无法正常工作。任何帮助将不胜感激。

我试图将数据库中的一些数据作为JSON返回。

这是我的代码：

<?php

$db         = "signup";
$servername = "localhost";
$username   = "username";
$password   = "password";
$fetch = ("SELECT * FROM bookings"); 
$return_arr = array();

$conn = new mysqli($servername, $username, $password, $db); // working 
$query_result = mysqli_query($conn, $fetch);
mysqli_select_db("bookings",$db);

while ($row = mysqli_fetch_array($query_result, MYSQL_ASSOC)) {

    $return_arr[] = array(
        'TIMESLOT' => $row['TIMESLOT'],
        'USERPIN' => $row['USERPIN'],
        'PLAYERS' => $row['PLAYERS'],
        'ACCEPTED' => $row['ACCEPTED'],
        'EMAIL' =>  $row['EMAIL'],
        'TOTAL BOOKINGS' => $row['TOTALBOOKINGS'],
        'BOOKING ID' => $row['BOOKINGID'],
        'LAST BOOKING' => $row['LASTBOOKING'],
        'DATE' => $row['DATE']
    );
}

$user = json_encode($return_arr);

echo "<pre>";
print_r($return_arr);
echo "</pre>";     

mysqli_close($conn);

?>

现在我从数据库中获取此信息：

Array
(
    [0] => Array
        (
            [TIMESLOT] => 12:00-14:00
            [USERPIN] => 25225649
            [PLAYERS] => 4
            [ACCEPTED] => 0
            [EMAIL] => email@address.com
            [TOTAL BOOKINGS] => 0
            [BOOKING ID] => 1
            [LAST BOOKING] => 0000,00,00
            [DATE] => 0000,00,00
        )
)

Answer 1

我会将$ tclsh 2patt.tcl file foo4 foo4 #98980更改为：

while

看到我在//... $conn = new mysqli($servername, $username, $password, $db); // working $query_result = mysqli_query($conn, "SELECT * FROM bookings"); //... // Option 1 : Change MYSQL_ASSOC to MYSQLI_ASSOC while ($row = mysqli_fetch_array($query_result, MYSQLI_ASSOC)) { // Option 2 : Change to mysqli_fetch_assoc // while ($row = mysqli_fetch_assoc($query_result)) { // Debug echo "<pre>"; print_r($row); echo "</pre>"; $return_arr[] = array($row['BOOKINGID'], $row['USERPIN'], $row['TIMESLOT'], $row['PLAYERS'], $row['ACCEPTED'], $row['EMAIL'], $row['TOTALBOOKINGS'], $row['LASTBOOKING'], $row['DATE'] ); }添加了括号，否则您将更改数组的值，而您只会获得最后一个。

之后删除＆＃34; [0]＆＃34;来自$return_arr。这样你就可以将整个数组编码为json。

Answer 2

试试此代码

$conn = new mysqli($servername, $username, $password, $db); // working 

$sql = ""; // your query
$result = $conn->query($sql);

while($row = $result->fetch_array()) {
    // Create results array
    $rows[] = array(
                    'BOOKINGID' => $row['BOOKINGID'],
                    'USERPIN' => $row['USERPIN'],
                    // and go on....
                );                              
}

mysqli_close($conn);

echo json_encode($rows[0]);

php不会从db返回json数据

2 个答案: