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php - 使用 AJAX 将 php 信息加载到 html 文档中

转载 作者:搜寻专家 更新时间:2023-10-31 21:20:04 25 4
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我对使用 AJAX 很陌生,在用户单击 svg 图像的一部分后,使用它从 php 处理程序文件将数据加载到 html div 元素时遇到问题。

HTML 代码——在 php 发送数据库信息后由 ajax 更新

<div  class="col-md-5 col-sm-5 exercises">
<h2 style="text-align: center;">Exercises</h2>
<div class="exercise-list">
<?php //$exercises->getExercises($result); ?>
<div class='media'><h3 id="exercise-list">Pick a muscle</h3>
</div>
</div>
</div>

ExerciseHandler.php -- 无法弄清楚 $_GET 中的内容,还是我应该使用不同的方法?

<?php
require_once "Exercises.class.php";
include 'dbconnect.php';

$connect = new mysqli($servername, $username, $password, $dbname);

$sql = "SELECT Muscle, Exercise, ImgResource, Descrip FROM exercises";
$result = $connect->query($sql);

if(isset( $_GET[''] )) {
$exercises = new Exercises();
$exercises->getExercises($result);
}
?>

练习.class.php

class Exercises {
public function getExercises($result) {
if ($result->num_rows > 0) {
// output data of each row
while ($row = $result->fetch_assoc()) {
echo "<div class='media'>";
echo "<div class='media-object pull-left'>";
echo "<a href='#'><img src='".$row["ImgResource"]."' class='img-responsive' alt='curl'></a>";
echo "</div>";
echo "<div class='media-body'>";
echo "<h4 class='media-heading'><a href='#'>".$row["Exercise"]."</a></h4>";
echo "</div>";
echo "</div>";
}
}
}
}
?>

loadExercises.js

function getExercises(clicked_muscle){
var muscle = clicked_muscle;
alert("You selected" + muscle);

if (window.XMLHttpRequest) {
//modern browsers
ajaxRequest = new XMLHttpRequest();
} else {
// code for old IE browsers
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}

ajaxRequest.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("exercise-list").innerHTML=this.responseText;
};

}

ajaxRequest.open("GET", "ExerciseHandler.php", true);
ajaxRequest.send();


}

获取练习函数。肌肉的事实之一

 public function getExercises($result) {        
if ($result->num_rows > 0) {
// output data of each row
while ($row = $result->fetch_assoc()) {
echo "<div class='media'>";
echo "<div class='media-object pull-left'>";
echo "<a href='#'><img src='".$row["ImgResource"]."' class='img-responsive' alt='curl'></a>";
echo "</div>";
echo "<div class='media-body'>";
echo "<h4 class='media-heading'><a href='#'>".$row["Exercise"]."</a></h4>";
echo "</div>";
echo "</div>";
}
}

}

这是截图 Muscles facts

最佳答案

请检查下面的代码。

HTML 代码 我在这个文件中对选择选项做了一些更改,并为调用的 ajax 调用了 jquery 函数。

<div  class="col-md-5 col-sm-5 exercises">
<h2 style="text-align: center;">Exercises</h2>
<div class="exercise-list">
<select onchange="getExercises(this)">
<option value="1">Test</option>
<option value="2">Test2</option>
</select>
<?php //$exercises->getExercises($result); ?>
<div class='media'><h3 id="exercise-list">Pick a muscle</h3>
</div>
</div>
</div>

loadExercises.js 我在 GET 方法中传递了数据。

function getExercises(clicked_muscle){
var muscle = set.value;;
alert("You selected" + muscle);
if (window.XMLHttpRequest) {
//modern browsers
ajaxRequest = new XMLHttpRequest();
} else {
// code for old IE browsers
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}

ajaxRequest.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("exercise-list").innerHTML=this.responseText;
};

}
ajaxRequest.open("GET", "ExerciseHandler.php?value="+muscle, true);
ajaxRequest.send();
}

ExerciseHandler.php

<?php
require_once "Exercises.class.php";
include 'dbconnect.php';

$connect = new mysqli($servername, $username, $password, $dbname);

$sql = "SELECT Muscle, Exercise, ImgResource, Descrip FROM exercises";
$result = $connect->query($sql);

if(isset($_GET) && !empty($_GET)) {
$exercises = new Exercises();
$exercises->getExercises($result);
}
?>

关于php - 使用 AJAX 将 php 信息加载到 html 文档中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54340787/

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