Php 如何通过单击按钮调用类函数
我有一个PHP类,它有一个插入数据的方法。创建类的实例后,我不确定如何调用该方法。我已经了解了如何完成调用方法,但是没有一个涉及调用类实例的方法 我的PHP是:Php 如何通过单击按钮调用类函数,php,Php,我有一个PHP类,它有一个插入数据的方法。创建类的实例后,我不确定如何调用该方法。我已经了解了如何完成调用方法,但是没有一个涉及调用类实例的方法 我的PHP是: <?php $servername = "localhost"; $username = "kevin"; $password = "ally"; $database = "test"; // Create connection $conn = new mysqli($servername, $username, $passw
<?php
$servername = "localhost";
$username = "kevin";
$password = "ally";
$database = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['btnInsert']) && ($_POST['btnInsert'] == "Insert"))
{
$Dog = new Dog($_POST['txtName'], $_POST['txtSize'], $_POST['txtColor'], $_POST['txttype'], $_POST['txtDescription']);
}
class Dog
{
public $name = "Dog Name";
public $size = 0;
public $color = "255:255:255";
public $typeName = "type Name";
public $typeDescription = "type Description";
public function Dog($name, $size, $color, $type, $description){
$this->name = $name;
$this->size = $size;
$this->color = $color;
$this->typeName = $type;
$this->typeDescription = $description;
}
public function InsertDog(){
$query = "SELECT Id from tbl_type WHERE Name = $typeName";
$result = mysqli_query($conn, "INSERT INTO tbl_type($this->typeName, $this->$typeDescription)") or die("Query fail: " . mysqli_error());
$result2 = mysqli_query($conn, "INSERT INTO tbl_Dog($this->$name, $this->$size, $this->$color, $query)") or die("Query fail: " . mysqli_error());
}
public function UpdateDog(){
}
}
?>
我的表格:
<form action="index.php" method="post">
Dog Name:<br />
<input name="txtName" type="text" /><br />
<br />
Size:<br />
<input name="txtSize" type="text" /><br />
<br />
Color:<br />
<input name="txtColor" type="text" /><br />
<br />
type Name:<br />
<input name="txttype" type="text" /><br />
<br />
type Description:<br />
<input name="txtDescription" style="width: 419px; height: 125px" type="text" /><br />
<br />
<input name="btnInsert" type="submit" value="Insert" />
</form>
狗名:
大小:
颜色:
类型名称:
类型说明:
如何为$Dog
实例调用InsertDog()
方法?只需使用:
...
$Dog = new Dog($_POST['txtName'], $_POST['txtSize'], $_POST['txtColor'], $_POST['txttype'], $_POST['txtDescription']);
$Dog->InsertDog();
:)您可以将所有代码放在一个文件(index.php)中,然后使用:
$Dog = new Dog($_POST['txtName'], $_POST['txtSize'], $_POST['txtColor'], $_POST['txttype'], $_POST['txtDescription']);
$Dog->InsertDog();
或者第二种解决方案是将ajax添加到代码中。所以链接jquery:
<script src="//code.jquery.com/jquery-1.11.2.min.js"></script>
<script src="//code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
编辑
到
并加上:
<script>
$( document ).ready(function() {
$( "#buttonId" ).click(function() {
$.ajax({
type: "POST",
url: "index.php",
data: { txtName: "DogsName", txtSize: "DogSize", txtColor: "color", txttype: "type", txtDescription: "dscr"v}
})
.done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
});
</script>
$(文档).ready(函数(){
$(“#按钮ID”)。单击(函数(){
$.ajax({
类型:“POST”,
url:“index.php”,
数据:{txtName:“DogsName”,txtSize:“DogSize”,txtColor:“color”,txttype:“type”,txtDescription:“dscr”v}
})
.done(函数(msg){
警报(“保存的数据:“+msg”);
});
});
});
指向index.php的url必须是server/index.php,所以如果使用localhost,它必须是
我没有测试它,但它应该可以工作
<input name="btnInsert" type="submit" value="Insert" id="buttonId" />
<script>
$( document ).ready(function() {
$( "#buttonId" ).click(function() {
$.ajax({
type: "POST",
url: "index.php",
data: { txtName: "DogsName", txtSize: "DogSize", txtColor: "color", txttype: "type", txtDescription: "dscr"v}
})
.done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
});
</script>