问:我无法找到我的代码没有工作的原因。Ajax/PHP:返回值
的index.php
while($row = mysqli_fetch_array($result))
{
$Tri_CAPTION=$row['BOOKCAPTION'];
$Tri_IMAGE=$row['BOOKIMAGE'];
$BOOKT=$row['BOOKTITLE'];
$BOOKID=$row['BOOKID'];
$html = $Tri_IMAGE;
$doc = new DOMDocument();
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);
$src = $xpath->evaluate("string(//img/@src)"); # "/images/image.jpg"
echo "<div class=\"mybooks\">
<a href=\"".$src."\" data-title=\"$BOOKT\" data-lightbox=\"example-3\" alt=\"Home\">
<img src=\"".$src."\" width=\"150\" height=\"150\"/>
</a>
<br />
Title:<strong>
".$BOOKT."</strong>
<br />
<a onClick=\"forusedata($BOOKID);\" href=\"#\">View Book</a>
</div>";
}
?>
阿贾克斯:
<script>
function forusedata(newid)
{
$.post('insert_home.php',{newid:newid}).done(function(data){
alert(data);
});
$('#Page1').dialog('open');
}
</script>
insert_home.php
else if(isset($_POST['newid']))
{
$newid = ($_POST['newid']);
$result = mysqli_query($con,"SELECT BOOKCAPTION FROM book WHERE BOOKID='$newid'");
while($row = mysqli_fetch_array($result)){
echo $row['BOOKCAPTION'];
}
mysqli_close($con);
}
让我们从基础开始。应该发生什么,实际发生了什么?将任何错误消息添加到您的问题。检查服务器错误日志并添加您在其中找到的任何相关错误。 – 2014-01-22 22:43:39
它shoukd输出html标签,因为html标签是BOOKCAPTION的内容,会发生什么,它不会输出任何东西 – user3196424