我这么写PHP后台接口为什么只能返回数据库一条数据 我要的是返回跟uphonenumber相匹配的所有数据

Question

<?php
    header("Content-Type:application/json;charset=UTF-8");
    $uphonenumber = $_POST['phonenumber'];//post获得手机号码表单值
    $json = '';
    $data = array();
    include('init.php');//链接数据库
    
    //$sql = "SELECT * FROM yffice_project where uphonenumber = $uphonenumber";
    $sql = "SELECT * FROM yffice_project where uphonenumber = $uphonenumber";
    mysql_query("set names utf8");
    $result = mysql_query($sql);//执行sql
    if($result){
    //echo "查询成功";
    while ($row = mysql_fetch_array($result))
    {
    $data["uid"]=$row["uid"];
    $data["pcontent"]=$row["pcontent"];
    $data["src"]=explode(",",$row["ppic"]);
    $data["pdate"]=$row["pdate"];
    $data["pweek"]=$row["pweek"];
    $data["pweather"]=$row["pweather"];
    }
    $json=json_encode($data);
    echo $json;
    }else{
    echo "查询失败";
    }
    mysql_close();//关闭数据库
?>

Answer 1

你的while循环里的赋值有问题，一直在重复为同一个键名赋值。
正确的应该是：

$data[] = [
    'uid'      => $row["uid"],
    'pcontent' => $row["pcontent"],
    'src'      => explode(",",$row["ppic"]),
    'pdate'    => $row["pdate"],
    'pweek'    => $row["pweek"],
    'pweather' => $row["pweather"]
];

这种形式；

Answer 2

循环写法有问题。你这样写，当前数据都会把之前的数据覆盖掉，最后只返回最后一条数据