情况非常简单，发现这很奇怪，为什么我无法在我的组合框（下拉菜单）中的每个结果后插入图像。

我正在尝试在此处插入$image！

       <select name = 'pSelect' id = 'pSelect'>       
      <?php
        $result = mysql_query
        ("                    
            SELECT ID, Project, Projectnummer, Klant
            FROM tblproject
            WHERE Status = '1'
            ORDER BY Klant ASC
        ");
         $image = "<img src='images/status_groen.png' width='15' height='15' /> ";
        while($row1 = mysql_fetch_array($result))
           {
                $pID = $row1['ID'];
                echo "<option value='"" . $row1['ID'] . "'"";
                if (isset($_POST['pSelect']) &&  $row1['ID'] == $_POST['pSelect']) 
                { 
                echo " selected='selected'"; 
                } 
                echo ">" . $row1['Klant'] ." ". $row1['Project'] ." ". $row1['Projectnummer'] ." ".$image. "</option>";
                echo "<br/>'";
           }
        ?>
    </select>

数据输出后，它以红色显示图像语法，表明这可能是语法问题，但不一定。

源输出：

 <option value="202">3DNL reCAPTCHA toevoegen 13097/1 <img src='images/status_groen.png' width='15' height='15'/> </option><br/>