無法執行PHP文件中的Javascript代碼
[英]Unable to Execute Javascript Code In A PHP File
問題描述
我有一個將數據輸入到mysql數據庫中的php文件。 現在，我想在成功插入數據后顯示javascript消息，但是我無法執行javascript代碼。
下面是我的代碼。
<html> <head> <title></title> <script type="text/javascript"> function run(){ alert("Data Inserted Successfully"); } </script> </head> <body> <?php $servername = "localhost"; $username = "root"; $password = ""; $dbname = "datacentre"; $first=$_POST['firstname'];//this values comes from html file after submitting $last=$_POST['lastname']; $dept= $_POST['department']; $unit= $_POST['unit']; $request=$_POST['request']; $purpose=$_POST['purposebuttons']; $accessedby = $_POST['personbuttons']; $description=$_POST['description']; $accessdate = $_POST['date-time']; /* Get Current Date and Time for the bookking_time field */ $booking_time=new DateTime(); $booking_time = $booking_time -> format("Ymd H:i:s"); // Create connection $con = new mysqli($servername, $username, $password, $dbname , 3306); if ($con->connect_error) { die("Connection failed: " . $con->connect_error); } //mysql_select_db("$database", $con); $sql= "INSERT INTO data_centre_users (first_name,last_name,department, unit, request, purpose , accessed_by, description,booking_time,access_time) VALUES ('$first','$last','$dept', '$unit','$request','$purpose', '$accessedby' ,'$description', NOW() , '$accessdate')"; if ($con->query($sql) === TRUE) { /* Calling the javascript code */ echo '<script> run(); </scrit>'; } else { echo "Error: " . $sql . "<br>" . $con->error; } $con->close(); ?> </body> </html>
3 個解決方案
解決方案1
0 2015-09-07 10:03:23
腳本結束標記拼寫錯誤。 如下更新：
echo '<script> run(); </script>';
解決方案2
0 已采納 2015-09-07 10:04:11
檢查script拼寫
echo '<script> run(); </scrit>';
^^^^^^
這將是
echo '<script> run(); </script>';
解決方案3
0 2015-09-07 10:13:21
您拼錯了</scrit>它一定是</script> 。
您可以嘗試以下方法：
echo '<script> alert("Data Inserted Successfully"); </script>'
使用PHP $ _GET執行JavaScript代碼
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