Question

我需要一个答案，为什么我不能在代码中显示变量或将数据插入数据库中，首先是我需要插入，但是如果未插入，它不会显示错误代码，提交后，它只会刷新页面。我的插入人员有什么问题？或其他任何代码错误？

<?php 
$con=mysqli_connect("localhost","root","","patient");
?>
<html>
<title>Patient record</title>
<head>
<style>
.align{
    text-align:center;
}
</style>
</head>
<body>
<div class="align">
<form method="post">
<input name="patient" type="text" placeholder="Patient name" required/>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Edit</a>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Delete</a>
<br>
<br>
<input name="aphysician" type="text" placeholder="Admitting Physician" required/>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Edit</a>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Delete</a>
<br>
<br>
<input name="tphysician" type="text" placeholder="Treating Physician" required/>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Edit</a>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Delete</a>
<br>
<br>
Treatment:
<br>
<br>
<input name="time" type="text" placeholder="Date and time" required/>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Edit</a>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Delete</a>
<br>
<br>
<input name="result" type="text" placeholder="Result" required/>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Edit</a>&nbsp;&nbsp;<a href="edit.php?user="$_POST['id']">Delete</a>
<br>
<input type="submit" name="submit" value="Submit">
</form>
</div>
</body>

<?php
if(isset($POST['submit'])){
    $patient=$_POST['patient'];
    $aphysician=$_POST['aphysician'];
    $tphysician=$_POST['tphysician'];
    $time=$_POST['time'];
    $result=$_POST['result'];

    $q="INSERT INTO `details`(`patient`,`aphysician`,`tphysician`,`time`,`result`)VALUES
    ('".$patient."','".$aphysician."','".$tphysician."','".$time."','".$result."')";
    $r=mysqli_query($con,$q);
if($r){
        echo 'data inserted';
        echo'<br><br>';
        echo '"'.$patient.'" &nbsp;&nbsp; "'.$aphysician.'" &nbsp;&nbsp; "'.$tphyscician.'" &nbsp;&nbsp; "'.$time.'" &nbsp;&nbsp; "'.$result.'"';
        echo'<br>';

    }else{
        echo $q;
    }

}



?>
</html>

Answer 1

第1步。向表单标签添加操作

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

使您的表单提交到同一页面

第2步。提交后，您将使用

if($_SERVER["REQUEST_METHOD"] == "POST")

代替

    if(isset($POST['submit']))

为什么在插入和显示时出错，却没有插入和显示

1 个答案: