Question

这是我用于从MySQL获取数据的PHP URL。我需要制作mysqli_fetch_array代码，说明uid表中的填充app-data与uid表中的users相同，以获取app-data中所有行的数据{1}} uid向每个用户展示app-data表中的项目。

$host = "";
$user = "";
$pwd = "";
$db = ""; 

$con = mysqli_connect($host, $user, $pwd, $db);

if(mysqli_connect_errno($con)) {
    die("Failed to connect to MySQL: " . mysqli_connect_error());
} 

$sql = "SELECT * FROM app_data ORDER By id";
$result = mysqli_query($con, $sql);

$rows = array();

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    $rows[] = $row; 
}

mysqli_close($con);

echo json_encode($rows);

Answer 1

我认为这是正确答案：

<?php

$host = "";
$user = "";
$pwd = "";
$db = ""; 

$con = mysqli_connect($host, $user, $pwd, $db);

if(mysqli_connect_errno($con)) {
    die("Failed to connect to MySQL: " . mysqli_connect_error());
} 

$sql = "SELECT * FROM app_data WHERE u_id=".$_POST['posted_uid']." ORDER By id";
$result = mysqli_query($con, $sql);

$rows = array();

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    $rows[] = $row; 
}

mysqli_close($con);

echo json_encode($rows);

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android - php通过用户id获取mysql数据

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