在PHP中使用重音符号处理JSON对象

我试图从MySQL结果创建一个json对象，但没有得到我需要的结果。我不知道问题出在哪里。我在本地主机上执行，没有任何东西（白色窗口）

我想要这样的东西：

{
    arraytwo:{   0:{pid:"..",name:"..",year:"..",subject:".."},
                 1:{pid:"..",name:"..",year:"",subject:".."},
                 2:{pid:"..",name:"..",year:"..",subject:".."},
                  ...},
    success:1
}

//-------------我的代码在这行下面--------------------------

$response = array(); //array one (multi.)
$con = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_DATABASE);
$result = mysqli_query($con, "SELECT * FROM classroom") or die(mysqli_error());

if (mysqli_num_rows($result) > 0) {

    $response["arraytwo"] = array(); //array two

    while ($row = mysqli_fetch_array($result,MYSQLI_BOTH)) {

        $arraythree = array(); //array three

        $arraythree["pid"] = $row["pid"];
        $arraythree["name"] = $row["name"];
        $arraythree["year"] = $row["year"];
        $arraythree["subject"] = $row["subject"];

        array_push($response["arraytwo"], $arraythree); 
    }

    $response["success"] = 1;

    echo json_encode($response);
}

这是我的真实代码：

<?php

$response = array();

$con = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_DATABASE);

$result = mysqli_query($con, "SELECT *FROM products") or die(mysqli_error());

if (mysqli_num_rows($result) > 0) {

    $response["noticias"] = array();

    while ($row = mysqli_fetch_array($result,MYSQLI_BOTH)) {

        $noticia = array();

        $noticia["pid"] = $row["pid"];
        $noticia["nombre"] = $row["nombre"];
        $noticia["curso"] = $row["curso"];
        $noticia["asignatura"] = $row["asignatura"];
        $noticia["accion"] = $row["accion"];
        $noticia["descripcion"] = $row["descripcion"];
        $noticia["created_at"] = $row["created_at"];
        $noticia["updated_at"] = $row["updated_at"];


        array_push($response["noticias"], $noticia);    

        }
        echo json_encode($response);
        $response["success"] = 1;
        echo json_encode($response);

    } else {

        $response["success"] = 0;
        $response["message"] = "No notices found";
        echo json_encode($response);
    }
    ?>

---

您需要使用utf8_编码转换字符。将代码更改为：

    $noticia["pid"] = $row["pid"];
    $noticia["nombre"] = utf8_encode($row["nombre"]);
    $noticia["curso"] = $row["curso"];
    $noticia["asignatura"] = utf8_encode($row["asignatura"]);
    $noticia["accion"] = $row["accion"];
    $noticia["descripcion"] = utf8_encode($row["descripcion"]);
    $noticia["created_at"] = $row["created_at"];
    $noticia["updated_at"] = $row["updated_at"];

---

在phpmyadmin中执行SELECT*FROM checkool并告诉您看到了什么我看到了表中的3行。我不知道我的问题是什么我编辑我的答案，看一看我把我的真实代码，我想你可以看到所有信息的问题：/。我尝试了你的答案，我发现了同样的问题（没有）。我要输掉这场与php的战斗。非常感谢。你解决了我的问题，现在我真的很感谢你。我也理解我的问题。小心点，阿德里安。@SalvaS，不客气，你什么时候可以投票表决；-）

    $noticia["pid"] = $row["pid"];
    $noticia["nombre"] = utf8_encode($row["nombre"]);
    $noticia["curso"] = $row["curso"];
    $noticia["asignatura"] = utf8_encode($row["asignatura"]);
    $noticia["accion"] = $row["accion"];
    $noticia["descripcion"] = utf8_encode($row["descripcion"]);
    $noticia["created_at"] = $row["created_at"];
    $noticia["updated_at"] = $row["updated_at"];