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[英]Oracle SQL request in PHP not returning anything but works in SQL Developer
[英]SQL request not returning anything when replacing a string by a variable
我正在尝试设置一个简单的函数来在重命名主文件夹时更新子文件夹的路径。
这是我的代码:
case 'rename':
$dirpath = rtrim(preg_replace('~/+~', '/', (str_replace('\\', '/', $_POST['old-dirpath']))), '/');
$new_dirpath = rtrim(preg_replace('~/+~', '/', (str_replace('\\', '/', $_POST['new-dirpath']))), '/');
$sql = "
UPDATE privatedir SET
dirpath = '{$new_dirpath}'
WHERE dirpath = '{$dirpath}'
";
$db->execute($sql);
$rs = $db->query("SELECT * FROM privatedir WHERE `dirpath` LIKE '%.$dirpath.%'");
$folders = $db->fetchAll($rs);
print_r ($folders); print_r($dirpath); die;
foreach ($folders as $folder){
echo $folder['dirpath']; die;
$folder_new_dirpath = str_replace($dirpath, $new_dirpath, $folder['dirpath']);
$sql = "
UPDATE privatedir SET
dirpath = '{$new_dirpath}'
WHERE dirpath = '{$folder['dirpath']}'
";
$db->execute($sql);
}
我的问题涉及到$rs = $db->query("SELECT * FROM privatedir WHERE
dirpath LIKE '%.$dirpath.%'");
即使我的数据库中有匹配的结果,它也不会返回任何信息,如果我使用print_r $folders
它只会向我输出一个空数组。
但是,如果我替换.$dirpath.
直接通过字符串(例如C:/wamp/www/gg/ftp/repository/folder1
),它将返回匹配的结果给我,而print_r $folders
将打印出对应的数组。
我究竟做错了什么 ?
像thi这样将变量放在引号中
$rs = $db->query("SELECT * FROM privatedir WHERE `dirpath` LIKE '%'".$dirpath."'%'");
事实是,您不需要周围的圆点,因为您不需要串联字符串。
这应该工作:
$rs = $db->query("SELECT * FROM privatedir WHERE `dirpath` LIKE '%$dirpath%'");
你也可以
$rs = $db->query("SELECT * FROM privatedir WHERE `dirpath` LIKE '%".$dirpath."%'");
如果您更喜欢它。
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