当我从数组$rawgendir=$row['gendir'];
中获取目录字符串时
scandir返回错误
"Warning: scandir( ALABAMA/CITIES ABBEVILLE - ALABAMA REVENUE ENFORCEMENT/ , ALABAMA/CITIES ABBEVILLE - ALABAMA REVENUE ENFORCEMENT/ ): The system cannot find the path specified. (code: 3)"
在var $ dir中明确定义路径,只是不适用于我想要创建的内容。在这种情况下,图像的搜索引擎
我认为问题是,我在scandir( ALABAMA/...)
得到一个幽灵空间而且我不知道如何。我已经在这工作了很长时间,因为我绝对不是程序员。
//$searchstring PULLED FROM _POST
$db_conx = (MYSQLi CONNECTION)
$sql="SELECT * FROM `main` WHERE CONVERT(`title` USING utf8) LIKE '%" . $SearchString . "%'";
$result = mysqli_query($db_conx, $sql);
//-create while loop and loop through result set
while($row=mysqli_fetch_array($result)){
$patchtitle=$row['title'];
$id=$row['id'];
$patchpath=$row['dir'];
$rawgendir=$row['gendir'];
echo $rawgendir;
这一行
echo $rawgendir;
打印出ALABAMA/CITIES ABBEVILLE - ALABAMA REVENUE ENFORCEMENT/
没有神秘的空间
// $dir = ("ALABAMA/CITIES ABBEVILLE - ALABAMA REVENUE ENFORCEMENT/");
$fileList = scandir($rawgendir);
//echo $dir;
//echo $rawgendir;
//-display the result o
echo '
<table width="864" height="220" border="2">
<tr>
<td width="214" height="152"><center>
<img src="'.$patchpath.' " width="220" height="200" />
</center></td>
<td width="632"<h3>Patch: '. $patchtitle . '</h3></td>
</tr>
</table> </br>
';
}