我试图使用ExtJS创建一个简单的表单并将值发送到数据库但它不会将我输入值的内容插入到我的表单中。我不知道为什么它没有输入数据到数据库我不知道我的代码有什么问题...任何人都可以帮助我实现我想要做的事感谢高级人员
PS即时使用Extjs 5
//我的Extjs表格
Ext.define('Grid.view.main.Form', {
extend: 'Ext.form.Panel',
xtype:'form',
title: 'User Form',
id:'myformpanel',
height: 120,
width: '100%',
autoScroll:true,
defaults: {
xtype: 'textfield',
labelAlign: 'top',
padding: 10
},
layout: {
type: 'hbox'
},
items: [{
fieldLabel: 'TEST1',
name: 'Test1'
},{
fieldLabel: 'TEST2',
name: 'Test2'
},{
fieldLabel: 'TEST3',
name: 'Test3'
},{
fieldLabel: 'TEST4',
name: 'Test4'
},{
fieldLabel: 'TEST5',
name: 'Test5'
},{
fieldLabel: 'TEST6',
name: 'Test6'
},{
fieldLabel: 'TEST7',
name: 'Test7'
},{
fieldLabel: 'TEST8',
name: 'Test8'
},{
fieldLabel: 'TEST9',
name: 'Test9'
},{
fieldLabel: 'TEST10',
name: 'Test10'
},{
fieldLabel: 'TEST11',
name: 'Test11'
},{
fieldLabel: 'TEST12',
name: 'Test12'
},
],
buttons: [{
text: 'login',
name:'submit',
handler: function(){
Ext.Ajax.request({
url: 'data/testform.php',
method: 'POST',
params: Ext.getCmp('myformpanel').getForm().getFieldValues(),
success: function(response){
Ext.Msg.alert('success ' + response.status);
},
failure: function(response){
Ext.Msg.alert('server-side failure with status code ' + response.status);
}
});
}
});
}
}]
// testform.php
<?php
require_once('database_connection.php');
if(isset($_POST['submit'])){
$Test1=$_POST['Test1'];
$Test2=$_POST['Test2'];
$Test3=$_POST['Test3'];
$Test4=$_POST['Test4'];
$Test5=$_POST['Test5'];
$Test6=$_POST['Test6'];
$Test7=$_POST['Test7'];
$Test8=$_POST['Test8'];
$Test9=$_POST['Test9'];
$Test10=$_POST['Tes10t'];
$Test11=$_POST['Test11'];
$Test12=$_POST['Test12'];
$query="INSERT INTO testexcel (Test1,Test2,Test3,Test4,Test5,Test6,Test7,Test8,Test9,Test10,Test11,Test12) VALUES ('$Test1','$Test2','$Test3','$Test4','$Test5','$Test6','$Test7','$Test8','$Test9','$Test10','$Test11','$Test12')";
}
?>
答案 0 :(得分：1)
我猜你的请求体是空的不是吗？尝试将表单数据插入到Ajax请求的params:配置中。 http://docs.sencha.com/extjs/4.2.0/#!/api/Ext.Ajax
Ext.Ajax.request({
url: 'data/testform.php',
method: 'POST',
params: this.ownerCt.ownerCt.getForm().getFieldValues(),
success: function(response){
Ext.Msg.alert('success ' + response.status);
},
failure: function(response){
Ext.Msg.alert('server-side failure with status code ' + response.status);
}
});
this.ownerCt.ownerCt.getForm()不是获取表单的最佳方式 - 虽然我无法在您的示例中找到另一种方法来访问它。如果您愿意，可以将表单设为id并与Ext.getCmp(id)合作。这看起来像params: Ext.getCmp('myformpanel123').getForm().getFieldValues()。
答案 1 :(得分：0)
感谢@Hown_爵士帮助我...我弄清楚为什么不将我的数据插入数据库的问题是什么...因为我的数据库配置错误所以这是我的代码
//我的extjs表格面板
Ext.define('Grid.view.main.Form', {
extend: 'Ext.form.Panel',
xtype:'form',
title: 'User Form',
id:'myformpanel',
height: 120,
width: '100%',
autoScroll:true,
defaults: {
xtype: 'textfield',
labelAlign: 'top',
padding: 10
},
layout: {
type: 'hbox'
},
items: [
{
fieldLabel: 'TEST1',
name: 'Test1'
},{
fieldLabel: 'TEST2',
name: 'Test2'
},{
fieldLabel: 'TEST3',
name: 'Test3'
},{
fieldLabel: 'TEST4',
name: 'Test4'
},{
fieldLabel: 'TEST5',
name: 'Test5'
},{
fieldLabel: 'TEST6',
name: 'Test6'
},{
fieldLabel: 'TEST7',
name: 'Test7'
},{
fieldLabel: 'TEST8',
name: 'Test8'
},{
fieldLabel: 'TEST9',
name: 'Test9'
},{
fieldLabel: 'TEST10',
name: 'Test10'
},{
fieldLabel: 'TEST11',
name: 'Test11'
},{
fieldLabel: 'TEST12',
name: 'Test12'
},
],
buttons: [{
text: 'login',
name:'submit',
handler: function()
{
Ext.Ajax.request({
url: 'data/testform.php',
method: 'POST',
params: Ext.getCmp('myformpanel').getForm().getFieldValues(),
success: function(response){
Ext.Msg.alert('success ' + response.status);
},
failure: function(response){
Ext.Msg.alert('server-side failure with status code ' + response.status);
}
});
}]
});
// my testform.php
<?php
require_once('database_connection.php');
$Test1= isset($_POST['Test1']) ? $_POST['Test1'] : NULL;
$Test2= isset($_POST['Test2']) ? $_POST['Test2'] : NULL;
$Test3= isset($_POST['Test3']) ? $_POST['Test3'] : NULL;
$Test4= isset($_POST['Test4']) ? $_POST['Test4'] : NULL;
$Test5= isset($_POST['Test5']) ? $_POST['Test5'] : NULL;
$Test6= isset($_POST['Test6']) ? $_POST['Test6'] : NULL;
$Test7= isset($_POST['Test7']) ? $_POST['Test7'] : NULL;
$Test8= isset($_POST['Test8']) ? $_POST['Test8'] : NULL;
$Test9= isset($_POST['Test9']) ? $_POST['Test9'] : NULL;
$Test10= isset($_POST['Test10']) ? $_POST['Test10'] : NULL;
$Test11= isset($_POST['Test11']) ? $_POST['Test11'] : NULL;
$Test12= isset($_POST['Test12']) ? $_POST['Test12'] : NULL;
$query="INSERT INTO testexcel (Test1,Test2,Test3,Test4,Test5,Test6,Test7,Test8,Test9,Test10,Test11,Test12) VALUES ('$Test1','$Test2','$Test3','$Test4','$Test5','$Test6','$Test7','$Test8','$Test9','$Test10','$Test11','$Test12')";
mysql_query($query,$con);
?>
希望这段代码能帮助像我这样的初学者......
谢谢