表单提交代码
<form action="search.php" method="POST" id="form2">
<input type="text" name="Sr_no" required="true" placeholder="Enter Serial Number"><br>
<input type="submit" value="Check Status" name="ss">
</form>
我的search.php
<?php
$connect=mysqli_connect("localhost","root", "" ,"mydatabase");
$Sr_no = $_POST["Sr_no"];
$status_check = "SELECT verified FROM r_concession_forms WHERE id = '$Sr_id' LIMIT 1";
$s_check = $connect->query($status_check);
$s_value = $row = $s_check->fetch_assoc();
if($s_value["verified"] == "0")
{
echo "<script>
alert('Not Issued/Verified');
window.location.href='index.php';
</script>" ;
}
elseif($s_value["verified"] == "1"){
echo "<script>
alert('Issued and Verified');
window.location.href='index.php';
</script>";
}
else{
echo "<script>
alert('No Such Form Exists');
window.location.href='index.php';
</script>";
}
?>
输出: 我的输入首先检查数据库中是否存在该“id”
没有预期的警告框,但是URL显示search.php,但没有显示警告框
刷新页面后,提示框开始工作
页面的设计是用JavaScript
最佳答案
我认为您在查询中使用了之前 undefined variable $Sr_id。还要首先检查表格是否已发布。
刷新页面时代码工作正常,因为刷新页面时没有发布数据,因此查询返回空值并显示“不存在这样的表单”。但是当你发布页面时,我猜你在查询中给出的发布值是错误的。
<?php
if(!empty($_POST['ss'])) {
$connect=mysqli_connect("localhost","root", "" ,"mydatabase");
if(!empty($_POST['Sr_no'])) {
$Sr_no = $_POST["Sr_no"];
$status_check = 'select verified FROM r_concession_forms WHERE id = '.$Sr_no.' LIMIT 1';
$s_check = $connect->query($status_check);
$s_value = $row = $s_check->fetch_assoc();
if($s_value["verified"] == "0")
{
echo "<script>
alert('Not Issued/Verified');
window.location.href='index.php';
</script>" ;
}
elseif($s_value["verified"] == "1"){
echo "<script>
alert('Issued and Verified');
window.location.href='index.php';
</script>";
}
else{
echo "<script>
alert('No Such Form Exists');
window.location.href='index.php';
</script>";
}
}
}
?>
此外,我个人不喜欢在您提到的 PHP 之间使用脚本代码。尝试将代码分开以便更好地理解。
关于javascript - 提交表单后我必须刷新我的 php 页面以执行代码并显示输出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48487435/