jQuery populate items into a Select using jQuery ajax json, php

后端 未结 4 443
感动是毒
感动是毒 2020-12-14 05:05

I have a select field. I must fill with options taken from a mysql table.
Here is some little php code I have done using codeigniter framework

$idcateg =         


        
相关标签:
4条回答
  • 2020-12-14 05:26

    you could also just use $().load() and have your PHP code generate the <option> tags

      $return = "";
      while ($row = mysql_fetch_array($res)) {
        $value = $row['value'];
        $text = $row{'text'];
        $return .= "<option value='$value'>$text</option>\n";
      }
    print $return;
    }
    

    ...

    $('#select').load("<?=base_url()?>index.php/rubro/list_ajax/");
    
    0 讨论(0)
  • 2020-12-14 05:28

    Try the following Code.

    In Controller ---------

    public function AjaxTest() {
    
                $rollNumber = $this->input->post('rollNumber');
                $query = "";
    
                if($rollNumber !="")
                {
                   $query = $this->welcome_model->get_students();
                }
                else
                {
                   $query = $this->welcome_model->get_students_informationByRoll($rollNumber);
                }
    
                $array = array($query);
                header('Content-Type: application/json', true);
                echo json_encode($array);
    
            }
    

    In View Add a Select Option

    <input type="text" id="txtSearchRoll" name="roll" value="" />
    <input type="button" name="btnSubmit" value="Search Students" onclick="return CheckAjaxCall();"/>
    
         <select id="myStudents">
                    <option>
                        --Select--
                    </option>
                </select>
    

    Now Scripts ----

    function CheckAjaxCall()
                {
                    $.ajax({
                        type:'POST',
                        url:'<?php echo base_url(); ?>welcome/AjaxTest',                    
                        dataType:'json',
                        data:{rollNumber: $('#txtSearchRoll').val()},                    
                        cache:false,
                        success:function(aData){ 
    
                            $('#myStudents').get(0).options.length = 0;
                            $('#myStudents').get(0).options[0] = new Option("--Select--", "0");        
    
                            $.each(aData, function(i,item) {
                            $('#myStudents').get(0).options[$('#myStudents').get(0).options.length] = new Option(item[i].Name, item[i].roll);
                                                                                                                // Display      Value
                        });
    
                        },
                        error:function(){alert("Connection Is Not Available");}
                    });
    
                    return false;
                }
    

    Enjoy the code....

    0 讨论(0)
  • 2020-12-14 05:30

    It's not much different.

    $idcateg = trim($this->input->post('idcategory'));
    $result = array();
    $id = mysql_real_escape_string($idcateg);
    $res = mysql_query("SELECT * FROM subcategories WHERE category = $id");
    while ($row = mysql_fetch_array($res)) {
      $result[] = array(
        'id' => $row['subcatid'],
        'desc' => $row['description'],
      );
    }
    echo json_encode($result);
    

    with:

    $.post("<?=base_url()?>index.php/rubro/list_ajax/", { 
      'idcategory' : idc },
      function(data) {
        var sel = $("#select");
        sel.empty();
        for (var i=0; i<data.length; i++) {
          sel.append('<option value="' + data[i].id + '">' + data[i].desc + '</option>');
        }
      }, "json");
    
    0 讨论(0)
  • 2020-12-14 05:36

    Yes. You want to pass back a JSON-encoded array of objects containing name/value pairs. Then you can create your select iteratively using these.

    $.post("<?=base_url()?>index.php/rubro/list_ajax/",
        {'idcategory' : idc },
        function(data){
            var select = $('#selectName').empty();
            $.each(data.values, function(i,item) {
                select.append( '<option value="'
                                     + item.id
                                     + '">'
                                     + item.name
                                     + '</option>' ); 
            });
        }, "json");
    
    0 讨论(0)
提交回复
热议问题