我有一个表,它的第三列是复选框,它的前两列来自数据库我想发送检查的单词的 wordID:
$result = mysql_query("SELECT * FROM words");
echo "<table border='1'>
<tr>
<th>word</th>
<th>meaning</th>
<th>checking</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['word'] . "</td>";
$idd= $row['id'] ;
echo "<td>". "<div class='hiding' style='display:none'>".$row['meaning']."</div>"."</td>";
echo "<td>";
echo "<input class=\"box\" name=\"$idd\" type=\"checkbox\" value=\"\"> ";
echo "</td>";
echo "</tr>";
}
echo "</table>";
在我的功能中,我有这个:
function feedback(){
var boxes = document.getElementsByClassName('box');
for(var j = 0; j < boxes.length; j++){
if(boxes[j].checked) {
assign=1;
}
else{
assign=0;
}
$.ajax({
url: "assigner.php",
type: "POST",
data: { wordid: wordid, assign: assign}
}).done(function( e ) {
/*alert( "word was saved" + e );*/
});
}
}
我不知道如何从第一部分获取 WordId 并在我所说的第二部分中使用它