我有一个变量在PHP,我需要它的值在我的JavaScript代码。我怎么能把我的变量从PHP到JavaScript?

我有这样的代码:

<?php
$val = $myService->getValue(); // Makes an API and database call

在同一页上,我有JavaScript代码,需要$val变量的值作为参数传递:

<script>
    myPlugin.start($val); // I tried this, but it didn't work
    <?php myPlugin.start($val); ?> // This didn't work either
    myPlugin.start(<?=$val?>); // This works sometimes, but sometimes it fails
</script>

当前回答

将数据转换为JSON 调用AJAX来接收JSON文件 将JSON转换为Javascript对象

例子:

步骤1

<?php

   $servername = "localhost";
   $username = "";
   $password = "";
   $dbname = "";
   $conn = new mysqli($servername, $username, $password, $dbname);

   if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
   } 

   $sql = "SELECT id, name, image FROM phone";
   $result = $conn->query($sql);

   while($row = $result->fetch_assoc()){ 
      $v[] = $row;    
   }

  echo json_encode($v);

  $conn->close();
?>

步骤2

function showUser(fnc) {
   var xhttp = new XMLHttpRequest();

   xhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
         // STEP 3    
         var p = JSON.parse(this.responseText);
      }
   }
}

其他回答

假设变量总是整数。在这种情况下,这样做更容易:

<?PHP
    $number = 4;

    echo '<script>';
    echo 'var number = ' . $number . ';';
    echo 'alert(number);';
    echo '</script>';
?>

输出:

<script>var number = 4;alert(number);</script>

假设你的变量不是整数,但如果你尝试上面的方法,你会得到这样的结果:

<script>var number = abcd;alert(number);</script>

但在JavaScript中,这是一个语法错误。

在PHP中,我们有一个函数json_encode,它将字符串编码为JSON对象。

<?PHP
    $number = 'abcd';

    echo '<script>';
    echo 'var number = ' . json_encode($number) . ';';
    echo 'alert(number);';
    echo '</script>';
?>

因为abcd在JSON中是“abcd”,它看起来像这样:

<script>var number = "abcd";alert(number);</script>

你可以对数组使用相同的方法:

<?PHP
    $details = [
    'name' => 'supun',
    'age' => 456,
    'weight' => '55'
    ];

    echo '<script>';
    echo 'var details = ' . json_encode($details) . ';';
    echo 'alert(details);';
    echo 'console.log(details);';
    echo '</script>';
?>

你的JavaScript代码看起来是这样的:

<script>var details = {"name":"supun","age":456,"weight":"55"};alert(details);console.log(details);</script>

控制台输出

诀窍是这样的:

下面是使用该变量的“PHP”: <?php $name = 'PHP变量'; 回声的< >脚本; 返回'var name = '。json_encode()美元。';; 回声的> < /脚本; ? > 现在你有了一个名为“name”的JavaScript变量,下面是使用该变量的JavaScript代码: <脚本> console.log(“我无处不在”+名字); > < /脚本

PHP

$fruits = array("apple" => "yellow", "strawberry" => "red", "kiwi" => "green");
<script>
    var color = <?php echo json_encode($fruits) ?>;
</script>
<script src="../yourexternal.js"></script>

JS (yourexternal.js)

alert("The apple color is" + color['apple'] + ", the strawberry color is " + color['strawberry'] + " and the kiwi color is " + color['kiwi'] + ".");

输出

苹果是黄色的,草莓是红色的,猕猴桃是红色的 颜色是绿色。

这是在2022年对我有用的,我使用这个解决方案来获取当前用户的电子邮件

我使用PHP创建了一个短代码,并将其添加到PHP .function:

function my_get_current_user_email(){
    $current_user = wp_get_current_user();
    $email = $current_user->user_email;
    return $email;  
} 
add_shortcode( 'get_email', 'my_get_current_user_email');

然后使用div来包装短代码:

 <div id="target-content" style="display: none;">
[get_email]
 </div>

最后,使用JavaScript访问Div的内容:

const databox = document.getElementById("target-content");
const dataContent = databox.textContent;
console.log(dataContent)

这工作完美地为我想要的,我希望它也会为你工作。

将数据转换为JSON 调用AJAX来接收JSON文件 将JSON转换为Javascript对象

例子:

步骤1

<?php

   $servername = "localhost";
   $username = "";
   $password = "";
   $dbname = "";
   $conn = new mysqli($servername, $username, $password, $dbname);

   if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
   } 

   $sql = "SELECT id, name, image FROM phone";
   $result = $conn->query($sql);

   while($row = $result->fetch_assoc()){ 
      $v[] = $row;    
   }

  echo json_encode($v);

  $conn->close();
?>

步骤2

function showUser(fnc) {
   var xhttp = new XMLHttpRequest();

   xhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
         // STEP 3    
         var p = JSON.parse(this.responseText);
      }
   }
}