這裏有兩個表格,即工具和tool_use。array_search函數不返回值
工具表看起來像這樣
id name tools_names quantity type
13 cutting player cutting playerA,cutting playerB,cutting playerC 3 engineer
12 REFLECTORS REFLECTORSA,REFLECTORSB 2 team
tool_use表看起來像這樣
id user_id type tools
8 siraj engineer cutting playerA,cutting playerB
7 siraj team REFLECTORSB
6 siraj team REFLECTORSA
我想除了顯示tools_names插入tool_use表同時插入而是整個tools_names正在顯示eventhough結果看起來像在桌子裏。我的控制
public function ajax_tools()
{
$data['tools']=$this->Tools_model->view_available_tools($_POST['type']);
foreach ($data['tools'] as $key=>$val) {
$data['toolss'][] =explode(',',$val['tools_names']);
}
$data['tools_names'] = $this->Tools_model->get_tool_names($_POST['type'])->result();
foreach ($data['tools_names'] as $row)
{
if (($key =array_search($row->tools,$data['toolss'])) !== false)
{
unset($data['toolss'][$key]);
$data['toolss'] = array_values($data['toolss']);
}
}
return $data['toolss'];
$this->load->view('ajax_tools',$data);
}
這裏是我的模型
public function view_available_tools($type)
{
$this->db->order_by('id','desc');
$this->db->where('status',1);
$this->db->where('type',$type);
$query=$this->db->get('tools');
return $query->result_array();
}
public function get_tool_names($type)
{
return $this->db->get_where('tool_use',array('type'=>$type));
}
這是我的看法
<div class="form-group">
<label for="type" class="control-label">Type:</label>
<select name="type" id="type" class="form-control" required>
<option value="">please select</option>
<option value="team" <?php echo set_select('type','team'); ?>>Team</option>
<option value="engineer" <?php echo set_select('type','engineer'); ?>>Engineer</option>
</select>
</div>
<div class="form-group ">
<label for="tools" class="control-label">Tools:</label>
<select name="tools[]" id="tools" multiple="multiple" required>
<option value="">please select</option>
</select>
</div>
<script>
$('#type').change(function(){
var type=$('#type').val();
var url='<?php echo base_url(); ?>tools/tools_control/ajax_tools';
$.post(url, {type:type}, function(data)
{
$('#tools').html(data);
});
});
</script>
請幫我解決我的問題
應該不會吧是「return $ data ['tools']」而不是「return $ data ['toolss']」? – deChristo
不要使用額外的字母來指示一些其他變量(工具與工具) - >總是使用自我解釋名稱(loadedTools與toolNames) - CleanCode –
它會成爲一個問題 –