我是PHP的初学者。我有一张表,它值取mysql。 share
有一个a href
字段,我还有一个文本框..所以我只单击共享,我想同时获得两个值...但是我的代码无法正常工作。它只能在表的第一行工作。我只单击第二行,但不传递文本框值
在这里检查我的代码...
$attchquery= mysql_query("select * from eduploads order by id desc limit 5") ; echo "
<table style='width:100%' class='table'>" ; echo "
<tr>" ; echo "
<th>No</th>
<th>Date</th>
<th>File Name</th>
<th>FileSize</th>
<th>Attachemnt Code</th>
<th>File Link</th>
<th>Share</th>" ; echo "</tr>" ; $count = 0 ; while ($row = mysql_fetch_array($attchquery)) { $count++ ; echo "
<tr>" ; echo "
<td>",$count,"</td>
<td>",$row['filedate'],"</td>
<td>", $row['filename'].".".$row['fileext'],"
</td>
<td>", $row['filesize'],"
</td>
<td>",$row['fileid'],"</td>
<td>", "
<a href='".$valueres[' settingsvalue '],$row['filelink ']."'>Download</a></td>
<td>", "
<a id='the-link' href='?id=".$row[' fileaddname ']."' class='btn btn-primary'>Share</a>
</td>" ; echo "</tr>" ; } echo "</table>" ; ?>
<form>
<input type="text" name="singleemail" id="singleemail" value="testvaluepossing">
</form>
<script>
document.getElementById('the-link').onclick = function(e) {
var singleemail = document.getElementById('singleemail').value;
document.location.href = this.href + '&singleemail=' + singleemail;
e.preventDefault();
return false;
}
</script>
这是如何工作的?请提供良好的解决方案或示例。