php - 显示带有数据库选择内容的警报

Question

我对 JavaScript 的相关内容非常陌生。

我有一组动态行和与这些行对应的列。在一列中，我有一个按钮。当我单击它时，它会根据 Posted Competence_ID 在另一个页面中显示 select 查询的结果。

查询工作正常，当我点击按钮时我得到了正确的结果。但是，我现在想做的是在单击按钮时在警报中显示该消息并停留在同一页面上而不是打开新选项卡..

这是显示我使用的表格的相关 HTML 代码:

echo "<table border='1' id='mycompstable' class='sortablee' cellpadding='0' cellspacing='0'>";
echo "<tr><th>ID</th><th>Competence Group</th><th>Competence Class</th><th>Competence Description</th><th>Own Evaluation</th><th>Manager's evaluation from last year</th><th>Target levels</th><th>Gap</th><th>Action</th><th class='unsortable'>Action ready target </th></tr>";

foreach($descs as $compi){  
    echo "<tr>";
        echo "<td>".$compi['Competence_ID']."</td>";
        echo "<td><p style='text-align: center;'>".$compi['Competence_Group']."</p></td>";
        if(isset($compi['Competence_class'])){echo "<td>".$compi['Competence_class']."</td>";}else echo "<td><p style='text-align: center;'>-</p></td>";
        echo "<td>".$compi['Competence_Description']."</td>";
        echo "<td class='evaluation'>";
            echo "<select class='ownlevelselect' id='ownlevelselect-.".$compi['Competence_ID']."' name='level-".$compi['Competence_ID']."' >";
               if (isset($compi['ownlevel']) && $compi['ownlevel']!= '' && !empty($compi['ownlevel']) && $compi['ownlevel']!= 0) {
                  echo "<option selected value='".$compi['ownlevel']."' selected='selected'>".$compi['ownlevel']."</option>";

                }
                echo "<option value='' >--</option>";
                echo "<option value='1'>1</option><option value='2'>2</option><option value='3'>3</option><option value='4'>4</option><option value='5'>5</option>";
            echo "</select>";
            echo $compi['ownlevel'];


            echo '<a test="'.$compi['Competence_ID'].'" onClick="showLevels('.$compi['Competence_ID'].');" target="_blank" href="'.INDEX.'?categ='.$_GET['categ'].'&action='.$_GET['action'].'&subaction=viewlevels'.'&levels='.$compi['Competence_ID'].'">';
            echo '<img class="linkki" src="'.KUVAT.'paivita.gif" alt="'._("tiedot").'" title="'._("Click to view the description of each level?").'"/></a>';
        echo "</td>";

这是获取数据的代码:

    function fetchlevels($Competence_id){
    $this->query="SELECT * FROM levels WHERE comp_id=".$_REQUEST['levels'];
    $tulos=$this->suoritaKysely();
    return $tulos;
}

这是我想在消息中显示的页面:

$levels=$this->levels;
$comp=$this->compdesc;
echo "Levels explanation for the competence:".$comp['Competence_Description']."<br>";

echo "Level 1 =  ".$levels[0]['lvl1'];
echo "<br>"."level 2 = ".$levels[0]['lvl2'];
echo "<br>"."level 3 = ".$levels[0]['lvl3'];
echo "<br>"."level 4 = ".$levels[0]['lvl4'];
echo "<br>"."level 5 = ".$levels[0]['lvl5'];
echo "<br><br>";
echo '<input type="button" value="close" window onclick="window.close();">';



?>

非常感谢任何形式的帮助

Answer 1

这是jsfiddle中的Ajax模拟 http://jsfiddle.net/ncubica/Umxjb/

HTML

<i style='display:none' id="loadingPopup">Loading</i>
<table>
    <tr>
        <td data-id="td1"> row 1</td>
    </tr>        
   <tr>        
        <td data-id="td2"> row 2</td>
    </tr>
    <tr>
        <td data-id="td3"> row 3</td>
    </tr>        
</table>

javascript

$("table").on("click", function(event){
    var $target = $(event.target); 
    if($target.is("td")){
        var id = $target.attr("data-id");
        makeAjax(id);
    }
});

//simulating ajax

function makeAjax(id){
    //you will have to use ajax an retrieve you json format
    //i will simulate ajax only
    $("#loadingPopup").show();
    var _json = { id : id, value : "some value", description : "some description"};
    setTimeout(function(){response(_json)}, 1000);
}

function response(_json){
    $("#loadingPopup").hide();    
    alert(_json.id + " - " + _json.value);
}

CSS

table{font-family:arial; font-size:12px; width:100%}
table tr td{border-bottom:1px solid #CCC; padding:10px;}