Question

以下是我正在使用的代码，当我从数据库中选择值时，它们会被提交到数据库，但是如果我输入的内容不在数据库中并且我想要它提交，那么它就不会被PHP提交或回显。 Sombody请帮助我。

谢谢。

 <?php
//Jason File
      #Include the connect.php file
      include('db_connect2.php');
     //get county of selected district
     $query = "SELECT * FROM primary_schools ";

      $result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
      while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
          $customers[] = array(
              'Emis_No' => $row['Emis_No'],
              'District' => $row['District'],
              'County' => $row['County'],
              'Subcounty' => $row['Subcounty'],
              'Parish' => $row['Parish'],
              'School' => $row['School']
               );
      }

      echo json_encode($customers);
    ?>

//Script
<script type=”text/javascript”>
$(document).ready(function () {
        //start EMIS code
            var customersSourcel =
        {
            datatype: "json",
            datafields: [
                { name: 'Emis_No'},
                { name: 'District'},
                { name: 'County'},
                { name: 'Subcounty'},
                { name: 'Parish'},
                { name: 'School'}
            ],
            url: 'includes/emis.php',
            cache: false,
            async: false
        };

        var customersAdapterl = new $.jqx.dataAdapter(customersSourcel);

        $("#emis_no").jqxComboBox(
        {
            source: customersAdapterl,

            width: 200,
            height: 25,
            promptText: "emis",
            displayMember: 'Emis_No',
            valueMember: 'Emis_No'
        });

        $("#emis_no").bind('select', function(event) 
        {
            if (event.args)
            {
                var index = $("#emis_no").jqxComboBox('selectedIndex');     
                if (index != -1)
                {
                    var record = customersAdapterl.records[index];
                    document.form1.district.value = record.District;
                    $("#county").jqxComboBox({ disabled: false});
                    document.form1.county.value = record.County;
                    $("#sub_county").jqxComboBox({ disabled: false});
                    document.form1.sub_county.value = record.Subcounty;
                    $("#parish").jqxComboBox({ disabled: false});
                    document.form1.parish.value = record.Parish;
                    $("#school").jqxComboBox({ disabled: false});
                    document.form1.school.value = record.School;
                }
            }
        });

Answer 1

在PHP文件开头用$customers初始化$customers = array();数组，然后开始使用$customers[] = ...

将值推入其中

JQWidgets Combobox类型值未被回显或提交给数据库

1 个答案: