-1

我在应用 Codeigniter 期间遇到了一些问题——创建了一个类似的函数

function searchUnivtab() {
        $country = $this->input->post('countryKey');
        $state = $this->input->post('stateKey');
        $level = $this->input->post('level');
        $degType = $this->input->post('degType');       
        $country = str_replace('%20', ' ', $country);
        $state = str_replace('%20', ' ', $state);
        $degType = explode('~', $degType);
        $data = @$this->get->getSearchedUniversityTab($country, $state, $level, $degType[1]);
        $html = '';
        $i = 0;
        foreach($data as $d)
        {
            $html .= '<option value="'.$d['name'].'">'.$d['name'].'</option>';
        }
        echo $html; die;
    }

错误为: A PHP Error was encountered

Severity: Warning

Message: Invalid argument supplied for foreach(),在线Line Number: 270_foreach

与上述代码相关的任何帮助?

4

3 回答 3

1

删除以下行中的 @ 以查看它是否产生任何错误,因此请替换

$data = @$this->get->getSearchedUniversityTab($country, $state, $level, $degType[1]);

$data = $this->get->getSearchedUniversityTab($country, $state, $level, $degType[1]);
于 2012-11-19T05:28:08.323 回答
0
if(is_array($data)) {
    foreach($data as $d){
        $html .= '<option value="'.$d['name'].'">'.$d['name'].'</option>';
    }
} else {
    echo 'data is not an array.';
}
于 2012-11-19T05:36:23.313 回答
-1
function getSearchUniversity() {
  country = jQuery('[name=countryKey]').val();
  state = jQuery('[name=stateKey]').val();
  level= jQuery('[name=level]').val();
  degType= jQuery('[name=degType]').val();
  jQuery.ajax({
    type: "POST",
    url:baseurl+ 'welcome/searchUnivtab', 
    cache: false,
    data: {countryKey: country, stateKey: state, level: level, degType: degType},
    error: function()
    {
      //notify('Error: Your request could be processed.. try again..!!');
    },
    success: function(html)
    {
      jQuery('[name=universityList]').html(html);
      return false;
    }
  }); 
}
于 2012-11-20T07:57:18.497 回答