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Php 从MySQL数据库获取数据时未定义属性_Php_Mysql_Sql_Select - Fatal编程技术网

Php 从MySQL数据库获取数据时未定义属性

Php 从MySQL数据库获取数据时未定义属性,php,mysql,sql,select,Php,Mysql,Sql,Select,我尝试使用以下脚本获取数据: public function getReligion() { $sql = 'select refReligionId, refReligionNameEN FROM ref_religion group by refReligionNameEN'; $this->selectSql($sql); $results = $this->getResult();

我尝试使用以下脚本获取数据:

   public function getReligion()
    {
        $sql = 'select refReligionId, refReligionNameEN FROM ref_religion group by refReligionNameEN';
        $this->selectSql($sql);             
        $results = $this->getResult();              
        $forms = '
        <select name="slcReligion" id="slcReligion" style="width: 204px">
            <option value="" selected>Select Religion</option>
        ';              
        foreach($results as $result)
        {
            $forms .= '<option value="'.$result->refReligionId.'">'.$result->refReligionNameID.'</option>';
        }
        $forms .= '</select>';              
        return $forms;
    }
我可以使用这些脚本获取数据

问题是php页面出现以下错误:

Undefined property: stdClass::$refReligionId
Undefined property: stdClass::$refReligionNameID

您正在选择
refReligionNameEN
,但正在对象引用中使用
refReligionNameID
EN
vs.
ID
)。

我已修复它,但它有相同的错误。。。我正在使用脚本获取另一个数据,如:GetOccidentity,getnationality。如果我删除get Nationality函数。这个功能做得很好。我必须做什么。。。
    public function getResult($getArray = false)
{
    $encode = json_encode($this->result);

    if($getArray == true)
        $array = true;
    else
        $array = false;                                 

    if($this->numResults == 1)
        $results = '['.$encode.']'; 
     else
        $results = $encode;

    $result = json_decode($results, $array);

    return $result;
}
Undefined property: stdClass::$refReligionId
Undefined property: stdClass::$refReligionNameID