我创建了一个包含 3 个字段的表，并在其中存储了值。我创建了一个可以执行不同操作的表单。其中之一是检索。当我单击检索按钮时，会显示表中请求的数据，但显示在不同的页面上。我想在同一页面中显示它。我想单独使用 PHP 来做到这一点。这可能吗？
<!DOCTYPE html>
<html>
<head>
<title>Untitled Document</title>
</head>
<body>
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'DB';
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_error())
{
die("couldn't connect" . $conn->connect_error());
}
$id = $_POST['Id'];
$name = $_POST['Name'];
$blood = $_POST['BloodGroup'];
if(isset($_POST['insert'])){
$insert = "Insert into ins(Id, name, BloodGroup) values ('$id','$name', '$blood')" ;
if($conn->query($insert) === TRUE) {
echo ("Input data entered successfully");
} else {
echo ("Input data failed to be entered" . $conn->error());
}
$conn->close();
} elseif(isset($_POST['update'])) {
$update = "update ins set Name='".$name."', BloodGroup='".$blood."' where Id='".$id."'";
mysql_query($update);
if($conn->query($update) === TRUE) {
echo ("Data updated successfully");
} else {
echo ("Data cant be updated" . $conn->error());
}
$conn->close();
} elseif(isset($_POST['delete'])) {
$id = $_POST['Id'];
$delete = "delete from ins where Id='".$id."'";
if($conn->query($delete) === TRUE) {
echo ("Data deleted successfully");
} else {
echo ("Data cant be updated" . $conn->error());
}
$conn->close();
}
else {
$id = $_POST['Id'];
$retrieve = "SELECT * FROM ins WHERE Id = ".$id;
if ($result=mysqli_query($conn,$retrieve))
{
while ($row=mysqli_fetch_row($result))
{
echo '<table>
<tr>
<td>Id</td>
<td>Name</td>
<td>Blood Group</td>
</tr>
<tr>
<td>'.$row[0].'</td>
<td>'.$row[1].'</td>
<td>'.$row[2].'</td>
</tr>
</table>';
//$row[0],$row[1],$row[2]';
}
mysqli_free_result($result);
}}
$conn->close();
?>
<h2>SELECT THE OPERATION YOU WANT TO PERFORM<h2>
<form method="post">
Id: <input type="text" name="Id" />
Name: <Input type="text" name="Name" />
BloodGroup: <input type="text" name="BloodGroup" /><br /><br />
<input type="submit" name="insert" value="Insert" />
<input type="submit" name="update" value="Update" />
<input type="submit" name="delete" value="Delete" />
<input type="submit" name="retrieve" value="retrieve" />
</form>
</body>
</html>
最佳答案
elseif(isset($_POST['retrieve']))
{
$id = $_POST['Id'];
$retrieve = "SELECT * FROM `ins` WHERE `Id` ='".$id."'";
if ($result=mysqli_query($conn,$retrieve))
{
while ($row=mysqli_fetch_row($result))
{
echo '<table>
<tr>
<td>Id</td>
<td>Name</td>
<td>Blood Group</td>
</tr>
<tr>
<td>'.$row[0].'</td>
<td>'.$row[1].'</td>
<td>'.$row[2].'</td>
</tr>
</table>';
}
mysqli_free_result($result);
}}
将其替换为 else 条件代码
关于php - 如何使用PHP在同一页面显示检索到的数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29649218/