Question

这是我的javascript代码：

<script type="text/javascript">
    $(document).ready(function(){
        $("#sepatu").click(function(e){
            e.preventDefault();
            var site_url = "<?php echo base_url()?>index.php/pelanggan/dyos_sepatu/";
            $("#content").load(site_url);
        })
    })
</script>

这就是我要加载的内容：

<select name="jenis" class="form-control">
        <?php
            foreach ($jenis->result() as $data){?>
            <option value="<?php echo $data->id_jenis_sepatu?>"><?php echo $data->nama_jenis?></option>
        <?php
            }
        ?>
    </select>

我在添加php脚本后出现问题：

<?php
        foreach ($jenis->result() as $data){?>
        <option value="<?php echo $data->id_jenis_sepatu?>"><?php echo $data->nama_jenis?></option>
    <?php
        }
    ?>

它不会加载。当我删除该脚本时，它会加载。这是我的控制器：

public function dyos_sepatu(){
        $data['jenis'] = $this->db->query("SELECT * FROM jenis_sepatu");
        $this->load->view('pelanggan/sepatu_dyos',true,$data);
    }

Answer 1

将数据发送到以下视图

public function dyos_sepatu(){
        $data['jenis'] = $this->db->query("SELECT * FROM jenis_sepatu")->result();
        $this->load->view('pelanggan/sepatu_dyos',true,$data);
    }

然后在$jenis->result()视图中$jenis。尝试一次

喜欢这个......

<?php
        foreach ($jenis as $data){?>
        <option value="<?php echo $data->id_jenis_sepatu;?>"><?php echo $data->nama_jenis;?></option>
    <?php
        }
    ?>

Answer 2

试试这个;

public function dyos_sepatu(){
    $data['jenis'] = $this->db->query("SELECT * FROM jenis_sepatu");
    echo $this->load->view('pelanggan/sepatu_dyos',true,$data); // note the echo here
}

需要将内容回显到页面jQuery .load()呈现页面。

Answer 3

我想你忘了在这个<?php echo $data->id_jenis_sepatu; ?>中终止php。还要检查选项标签

添加php脚本

3 个答案: