Alert() javascript 函數在 php 中不起作用

Question

我知道這是常見問題，但我嘗試使用：

1. script language='javascript'
2. 警報后將標題放在 else 中
3. script type='text/javascript'

我仍然沒有收到警報框，而其他部分執行得很好。

這是我的代碼：

  <?php

 /* header('Content-Type: application/json');
 $response = array(); */
 if (isset($_GET['sid'])){

$con = mysqli_connect("localhost", "root", "", "kaemiphk_greivance");

if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
 }

$checkdata=mysqli_query($con,"SELECT * FROM officer_master WHERE pf_no = '".$_GET['sid']."'");
$query_data=mysqli_num_rows($checkdata);
if ($query_data == 0) {

//echo alert "welcome";
    echo '<script type="text/javascript" src="https://code.jquery.com/jquery-1.11.0.min.js">';
    echo "alert('PF No. Does not exist. Please Contact Admin!!!');";
    echo '</script>';

 }
 else{
      header('Content-Type: application/json');
      $response = array();
      $select="SELECT m.officer_name,m.email,m.department,m.mobile_no,m.designation,n.quarter_no,n.address,n.colony,n.blueprint_quarter,n.type_of_quarter, n.area FROM officer_master m, quarter_master n WHERE n.pf_no='".$_GET['sid']."' AND m.pf_no = n.pf_no";
 $result = mysqli_query($con, $select);  //mysql_query($qry);
 while ($row = mysqli_fetch_assoc($result)) {
array_push($response, $row);
}
}
echo json_encode($response);  
} 

 ?>

我在這里錯過了什么。 謝謝

Answer 1

你把你的 js 文件搞混了。 在單獨的標簽中包含 jquery，然后包含您的腳本：

echo '<script type="text/javascript" src="https://code.jquery.com/jquery-1.11.0.min.js" ></script>';
echo '<script type="text/javascript">';
echo "alert('PF No. Does not exist. Please Contact Admin!!!');";
echo '</script>';

順便說一句，您不需要 jquery 來獲取簡單的警報，因為它是純 javascript。 如果不需要，盡量避免包含外部庫，你最終會得到一個臃腫的代碼。 用 php 打印 js 有點麻煩。 為什么不把它打印到你的 html 或 js 文件中？

Answer 2

具有 src 屬性的腳本標簽內的 Javascript 不會被執行，您必須在 jquery 標簽之后創建第二個腳本標簽。

<?php

/* header('Content-Type: application/json');
$response = array(); */
if (isset($_GET['sid'])){

   $con = mysqli_connect("localhost", "root", "", "kaemiphk_greivance");

   if (mysqli_connect_errno()){
     echo "Failed to connect to MySQL: " . mysqli_connect_error();
   }

   $checkdata=mysqli_query($con,"SELECT * FROM officer_master WHERE pf_no = '".$_GET['sid']."'");

   $query_data = mysqli_num_rows($checkdata);
   if ($query_data == 0) {
//echo alert "welcome";
    echo '<script type="text/javascript" src="https://code.jquery.com/jquery-1.11.0.min.js">';
    echo '</script>';
    echo "<script>alert('PF No. Does not exist. Please Contact Admin!!!');</script>";
   } else {
      header('Content-Type: application/json');
      $response = array();
      $select="SELECT m.officer_name,m.email,m.department,m.mobile_no,m.designation,n.quarter_no,n.address,n.colony,n.blueprint_quarter,n.type_of_quarter, n.area FROM officer_master m, quarter_master n WHERE n.pf_no='".$_GET['sid']."' AND m.pf_no = n.pf_no";
      $result = mysqli_query($con, $select);  //mysql_query($qry);
      while ($row = mysqli_fetch_assoc($result)) {
        array_push($response, $row);
      }
    }
  echo json_encode($response);  
  } 
}
 ?>