MySQL表填充一个dropown。选择后，将填充第二个下拉列表 [英] MySQL table populating one dropsown. Upon selection, A second dropdown is populated

问题描述

好吧，我已经在这里检查了网络和其他问题，我很难过。我试过了一个来自这里发布的问题的JavaScript解决方案，但我认为它不喜欢MySQL填充< option> s。我会复制所有我已经获得的JavaScript代码。

脚本：

 <脚本> $（$＃$ b $（'＃groups'）。on（'change'，function（）{
 var val = $（this）.val（）; 
 $ var $ =（'＃sub_groups'）; 
 $（'option'，sub）.filter（function（）{
 if（
 $（this）.attr（' （'data-group'）=== val || $（this）.attr（'data-group'）==='SHOW'
）{
 $（this）.show（）; 
} else {
 $（this）.hide（）; 
} 
}）; 
}）; 
 $（'＃groups'）。触发（'更改'）; 
}）; 
< / script> 
  

PHP第一个下拉列表：

 < select class =form-controlid =groups> 
<？php 
 $ sql =SELECT BoilerBrand FROM boilerbrands; 
 $ result = mysql_query（$ sql）; 
 while（$ row = mysql_fetch_array（$ result））{
 echo< option value ='。$ row ['ID']。'>。$ row ['BoilerBrand' ]。 < /选项> 中; 
} 
？> 
< / select> 
  

PHP第二个下拉列表

 < select class =form-controlid =sub_groups> 
< option data-group ='SHOW'value =0> Model< / option> 
<？php 
 $ sql =SELECT * FROM boilermodels; 
 $ result = mysql_query（$ sql）; 
 while $（$ row = mysql_fetch_array（$ result））{
 echo< option data-group ='。$ row ['BoilerBrand']。'value ='。$ row [ 'BoilerGC'] '> 中$行[ 'BoilerModel'] < /选项> 中。 
} 
？> 
< / select> 
  

任何帮助都将不胜感激！

感谢：）

解决方案

我通常这样做的方式不是隐藏/显示我删除/添加的选项。我相信如果你隐藏选项，那么选择输入仍然可以具有该值。

脚本：

 < script> 
 $（function（）{
<？php 
 $ sql =SELECT * FROM boilermodels; 
 $ result = mysql_query（$ sql）; 
 $ models = array（）; 
 while（$ row = mysql_fetch_array（$ result））{
 $ models [$ row ['BoilerBrand']] [] = $ row; 
} 
 / *应该看起来像
 $ models = []; 
 $ models [1] [] = ['ModelID'=>'1'，'BoilerBrand'=>'1'， 'BoilerModel'=>'240E'，'BoilerGC'=>'47 -777-77'，'BoilerImage'=> '47 -777-77.jpg']; 
 $ models [1] BoilerModel ='290D'，'BoilerGC'=>'11 -111-11'，'BoilerImage'=''ModelID'=>'3'，'BoilerBrand'=>'1' '=> '11 -111-11.jpg']; 
 $ models [2] [] = ['ModelID'=>'2'，'BoilerBrand'=>'2'，'BoilerModel '='250E'，'BoilerGC'=>'47 -777-77'，'BoilerImage'=> '47 -777-77.jpg']; 
 * / 
？ > 
 var _boilermodels ='<？php echo json_encode（$ models）;？>'; 
 var jsonBoilerModels = JSON.parse（_boilermodels）; 
 console.log（jsonBoilerModels）; $（$＃
 $（'＃groups'）。on（'change'，function（）{
 var $ this = $（this）; 
 var val = $ this.val （）; 
 var sub = $（'＃sub_groups'）; 
 sub.find（'option'）。remove（）; 
 var appendList = []; 
 
 $ .each（value）['BoilerGC']''>' ，value ['BoilerModel']，'< / option>'））; 
}）; 
 
 sub.append（appendList）; 
}）; 
 
 $（'＃groups'）。trigger（'change'）; 
}）; 
< / script> 
  

第一个下拉列表：

 < select class =form-controlid =groups> 
<？php 
 $ sql =SELECT ID，BoilerBrand FROM boilerbrands; 
 $ result = mysql_query（$ sql）; 
 while（$ row = mysql_fetch_array（$ result））{
 echo< option value ='。$ row ['ID']。'>。$ row ['BoilerBrand' ]。 < /选项> 中; 
} 
？> 
< / select> 
  

第二个下拉列表：

 < select class =form-controlid =sub_groups> 
< option value =>选择模型< / option> 
< / select> 
  

Ok, so I've checked the net and the other questions on here and I'm stumped. I've tried a javascript solution from a question posted on here but I think it's not liking MySQL populating the <option>s. I'll copy all the code I've got including the javascript I have.

SCRIPT:

<script>
  $(function() {
    $('#groups').on('change', function() {
      var val = $(this).val();
      var sub = $('#sub_groups');
      $('option', sub).filter(function() {
        if (
          $(this).attr('data-group') === val || $(this).attr('data-group') === 'SHOW'
        ) {
          $(this).show();
        } else {
          $(this).hide();
        }
      });
    });
    $('#groups').trigger('change');
  });
</script>

PHP 1st dropdown:

<select class="form-control" id="groups">
  <?php
        $sql = "SELECT BoilerBrand FROM boilerbrands";
        $result = mysql_query($sql);
        while ($row = mysql_fetch_array($result)) {
          echo "<option value='".$row['ID']."'>".$row['BoilerBrand']."</option>";
        }
        ?>
</select>

PHP 2nd dropdown

<select class="form-control" id="sub_groups">
  <option data-group='SHOW' value="0">Model</option>
  <?php
     $sql = "SELECT * FROM boilermodels";
     $result = mysql_query($sql);
     while ($row = mysql_fetch_array($result)) {
      echo "<option data-group='".$row['BoilerBrand']."' value='".$row['BoilerGC']."'>".$row['BoilerModel']."</option>";
     }
     ?>
</select>

Any help with this would be greatly appreciated!

Thanks :)

解决方案

The way I normally do this is instead of hiding/showing the options I remove/add them. I believe if you hide the options then the select input can still have that value.

SCRIPT:

<script>
    $(function(){
        <?php
        $sql = "SELECT * FROM boilermodels";
        $result = mysql_query($sql);
        $models = array();
        while ($row = mysql_fetch_array($result)) {
            $models[$row['BoilerBrand']][] = $row;
        }
        /* should look like
        $models = [];
        $models[1][] = ['ModelID'=>'1','BoilerBrand'=>'1','BoilerModel'=>'240E','BoilerGC'=>'47-777-77','BoilerImage'=>'47-777-77.jpg' ];
        $models[1][] = ['ModelID'=>'3','BoilerBrand'=>'1','BoilerModel'=>'290D','BoilerGC'=>'11-111-11','BoilerImage'=>'11-111-11.jpg' ];
        $models[2][]= ['ModelID'=>'2','BoilerBrand'=>'2','BoilerModel'=>'250E','BoilerGC'=>'47-777-77','BoilerImage'=>'47-777-77.jpg' ];
        */
        ?>
        var _boilermodels = '<?php echo json_encode($models); ?>';
        var jsonBoilerModels = JSON.parse(_boilermodels);
        console.log(jsonBoilerModels);

        $('#groups').on('change', function(){
            var $this = $(this);
            var val = $this.val();
            var sub = $('#sub_groups');
            sub.find('option').remove();
            var appendList = [];

            $.each(jsonBoilerModels[val],function(key,value){
                appendList.push('<option value="'.concat(value['BoilerGC'], '">', value['BoilerModel'], '</option>'));
            });

            sub.append(appendList);
        });

        $('#groups').trigger('change');
    });
</script>

1st Dropdown:

<select class="form-control" id="groups">
    <?php
        $sql = "SELECT ID ,BoilerBrand FROM boilerbrands";
        $result = mysql_query($sql);
        while ($row = mysql_fetch_array($result)) {
            echo "<option value='".$row['ID']."'>".$row['BoilerBrand']."</option>";
        }
    ?>
</select>

2nd Dropdown:

<select class="form-control" id="sub_groups">
    <option value="">Select A Model</option>
</select>

这篇关于MySQL表填充一个dropown。选择后，将填充第二个下拉列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！