php - 使用 PHP 从 MySQL 表中获取值并在 jQuery 函数中将结果显示为 "option"

Question

我面临以下问题:

1) 我使用 PHP MySQL 检索了“select->options”值，如下所示:

$fetch_sub_categories1 =    '{ label: "", value: "" },';
$fetch_sub_categories2 =    '{ label: "", value: "" },';
$fetch_sub_categories3 =    '{ label: "", value: "" },';
$fetch_sub_categories4 =    '{ label: "", value: "" },';


$num_of_returned_rows = mysqli_num_rows($get_sub_cats_query);        
if ($num_of_returned_rows>0)
{
for ($i=0;$i<$num_of_returned_rows;$i++)
{
     $row=mysqli_fetch_array($get_sub_cats_query);

     if ($row['category'] == 'Αρχικά Πιάτα'){
        if($i==$num_of_returned_rows) $fetch_sub_categories1 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" }';
        else                          $fetch_sub_categories1 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" },';
     }
     if ($row['category'] == 'Κυρίως Πιάτα') {
        if($i==$num_of_returned_rows) $fetch_sub_categories2 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" }';
        else                          $fetch_sub_categories2 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" },';
     }
     if ($row['category'] == 'Επιδόρπια')   {
        if($i==$num_of_returned_rows) $fetch_sub_categories3 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" }';
        else                          $fetch_sub_categories3 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" },';
     }
     if ($row['category'] == 'Ποτά')         {
        if($i==$num_of_returned_rows) $fetch_sub_categories4 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" }';
        else                          $fetch_sub_categories4 .= '{ label: ""'.$row['sub_category'].'"", value: ""'.$row['sub_category'].'"" },';
     }
}

}

2) 我有一个现成的文档监听器 jQuery 函数，我用它来显示上述结果，如下所示:

label: "Υποκατηγορία:",
                                name: "sub_category",
                                type:  "select",
                                options: [

                                        '<?php echo $fetch_sub_categories1; ?>'
                                ]

但是，浏览器会将整个变量 $fetch_sub_categories1 解释为字符串，并且不会显示变量中定义的选项。

结果如下:

任何建议都将受到高度赞赏。

Answer 1

删除 <?php ?> 周围的单引号标签:

label: "Υποκατηγορία:",
name: "sub_category",
type: "select",
options: [
    <?php echo $fetch_sub_categories1; ?>
]

它们就是 $fetch_sub_categories1 内容的原因被解释为字符串。