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php - 我如何在 PHP 编码中包含 ajax 以在同一页面中显示数据

转载 作者:行者123 更新时间:2023-11-30 22:45:43 25 4
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我想在同一页中显示从表中检索到的数据。我知道如果我使用 ajax,工作会变得更容易。但是我该怎么做呢?我对ajax一无所知。我现在的程序结合了html和php。该程序如下。我想要的是,当我单击按钮时,请求的操作和数据必须显示在同一页面中。

<!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 ($_SERVER["REQUEST_METHOD"] == "POST") {
if(isset($_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" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
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

$.ajax({
method: "POST",
url: "{php file url}",
data: { name: "John", location: "Boston" } // data list need to sent
})
.done(function( msg ) {
$("#div1").html(result); //#div1 is container element where you want to show output of ajax
});

更多详情请引用 http://api.jquery.com/jquery.ajax/

关于php - 我如何在 PHP 编码中包含 ajax 以在同一页面中显示数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29665940/

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