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Pages: [1] 2
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2
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General Category / EasyUI for jQuery / parse jason showing undefined error
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on: February 02, 2017, 02:02:16 AM
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here's my code:
function validateLogIn(){ var url = 'modules/data/get_loginuser.php'; alert(url);
$('#fm_login').form('submit',{ url: url, onSubmit: function(){ return $(this).form('validate'); }, success: function(result){ if (result.errorMsg){ $.messager.show({ title: 'Error', msg: result.errorMsg }); } else { alert(result); //result alert is: [{"user_id":"7818","user_name":"karogel","user_password":"827ccb0eea8a706c4c34a16891f84e7b","user_webpass":"827ccb0eea8a706c4c34a16891f84e7b","lastname":"ramiterre","firstname":"rogel","fullname":"rogel ramiterre","nameinitials":"rramiterre","jobtitle":"programmer","admin":"N","branch_name":"Guimba","active":"Y","online":"N","last_active":null,"dateadded":"2015-07-28 07:51:35","addedbyfk":"1","datemodified":"2015-07-28 09:28:47","lastuserfk":"1","is_sync":"0","datetime_sync":"0000-00-00 00:00:00"}] var vfirstname = (result['0'].firstname); alert(vfirstname); $.messager.alert('System Admin','Welcome'); // $('#dlg_login').dialog('close'); // close the dialog // redirect to main menu } } });
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4
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General Category / General Discussion / Re: How to handle onLoadSuccess if form load was used twice
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on: November 03, 2016, 07:39:10 PM
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var url_recom = 'modules/data/get_loanrecom.php?id='+row.maker_id+'&loanappid='+row.app_loanappid; $('#fm_application_proposal').form('load',url_recom);
$('#fm_application_proposal').form('load',row);
is the above code OK? i am trying to load the data from "url_recom" and "row" data (from datagrid).
if i will use onLoadSuccess, will it fire after the first form(load ? or after the last form(load or is therea way to comine the 2 data source into single form(load
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5
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General Category / General Discussion / How to handle onLoadSuccess if form load was used twice
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on: November 02, 2016, 01:15:03 AM
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I have this code: var flag; function loanproposalshow(){ var row = $('#dg_application_ci').datagrid('getSelected'); if (row){ // open dialog $('#dlg_application_proposal').dialog('open').dialog('setTitle','Loan Proposal');
$('#fm_application_proposal').form('clear'); var url_recom = 'modules/data/get_loanrecom.php?id='+row.maker_id+'&loanappid='+row.app_loanappid; $('#fm_application_proposal').form('load',url_recom);
$('#fm_application_proposal').form('load',row); $(function(){ $('#fm_application_proposal').form({ onLoadSuccess:function(){ flag = true; calc_recom_apa(); } }); }); url = 'modules/application/update_application.php?loanappid='+row.app_loanappid; } }
I am having a problem in calling calc_recom_apa function. It should initiate after the 2 form load and not from the first "form('load',url_recom)"
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11
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General Category / EasyUI for jQuery / Re: How to avoid onchange method to run during the initial loading of form data?
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on: August 11, 2016, 01:00:54 AM
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Sorry no luck! I want to use onchange only after loading the initial form data, here's my code:
<form id="fm_application" method="post" novalidate> <div> Charge 1: <input name="charge1" id="charge1" class="easyui-numberbox"></input></br> Charge 2: <input name="charge2" id="charge2" class="easyui-numberbox"></input></br> Charge 3: <input name="charge3" id="charge3" class="easyui-numberbox"></input></br> Total <input name="totalcharge" id="totalcharge" class="easyui-numberbox" readonly="true"></input></br> </div> </form>
<script> function editdata(){ var row = $('#dg_application').datagrid('getSelected'); if (row){ $('#dlg_application').dialog('open').dialog('setTitle','Edit Application'); var url_loancharge = 'modules/data/get_saveloancharge.php?appstatuscode=NewApp&loanappid='+row.app_loanappid; $('#fm_application').form('load',row); $.post(url_loancharge, function(data){ var data1 = $.parseJSON(data);//parse JSON var vtotalcharge = parseFloat(data1.app_charge1)+ parseFloat(data1.app_charge2)+ parseFloat(data1.app_charge3); $('#fm_application').form('load', $.extend({},data,{ charge1:data1.charge1, charge2:data1.charge2, charge3:data1.charge3, totalcharge:vtotalcharge })) }); } } $('#charge1').numberbox({ onChange: function(){ calc_totalcharge(); } }); $('#charge2').numberbox({ onChange: function(){ calc_totalcharge(); } }); $('#charge3').numberbox({ onChange: function(){ calc_totalcharge(); } }); function calc_totalcharge(){ var vcharge1 = parseFloat($('#charge1').numberbox('getValue')) || 0; var vcharge2 = parseFloat($('#charge2').numberbox('getValue')) || 0; var vcharge3 = parseFloat($('#charge3').numberbox('getValue')) || 0; var vtotalcharge = vcharge1 + vcharge2 + vcharge3; $('#totalcharge').numberbox('setValue', vtotalcharge); }
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15
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General Category / EasyUI for jQuery / Saving into multiple table
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on: November 01, 2015, 06:02:54 PM
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Please help...how to create save.php file with multiple table, say save to personalinfo.php and address.php
I am following or using this simple code for saving to single table:
<?php
$firstname = htmlspecialchars($_REQUEST['firstname']); $lastname = htmlspecialchars($_REQUEST['lastname']); $phone = htmlspecialchars($_REQUEST['phone']); $email = htmlspecialchars($_REQUEST['email']);
include '../conn.php';
$sql = "insert into users(firstname,lastname,phone,email) values('$firstname','$lastname','$phone','$email')"; $result = @mysql_query($sql); if ($result){ echo json_encode(array( 'id' => mysql_insert_id(), 'firstname' => $firstname, 'lastname' => $lastname, 'phone' => $phone, 'email' => $email )); } else { echo json_encode(array('errorMsg'=>'Some errors occured.')); } ?>
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