如何打印一个变量的所有值,以href从mysql数据库读取它,下面的变量是$src,值从数据库读取,如下所示
$file_query=mysqli_query($conn,"select * from tbl_taskimage where db_taskid='$id'")or die(mysqli_error($conn));
if(mysqli_num_rows($file_query)>0){
while($r=mysqli_fetch_array($file_query)){
$src=$r['db_image'];}
echo"<a href='download.php?src=$src'><img src='../img/download.png'></a>";
}当我把鼠标放在上面时,我希望结果是这样的。
....../download.php?src=xxx.png-yyy.jpg-rrr.jpeg
这有可能吗?怎么做?!!
发布于 2017-01-12 09:55:01
我想你是在找内爆:
<code>
$file_query=mysqli_query($conn,"select * from tbl_taskimage where db_taskid='$id'")or die(mysqli_error($conn));
$src = array();
if(mysqli_num_rows($file_query)>0){
while($r=mysqli_fetch_array($file_query)){
$src[]=$r['db_image'];}
}
echo '<a href="download.php?src='.implode('-',$src).'><img src="../img/download.png"></a>';
</code>https://stackoverflow.com/questions/41609882
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