我正在设计一个应用程序,在我的模态中遇到一个奇怪的错误,我无法编辑条目
下面是显示错误的模式:
<?php
include_once dirname(dirname(dirname(__FILE__))) . "/const.php";
include_once PHP_PATH . "/config1.php";
include_once CONFIG_PATH.'/modal/routemgmt/route_mgmt.php';
function sanitize($input) {
if (is_array($input))
return array_map('sanitize', $input);
else
return htmlspecialchars(trim($input));
}
// Sanitize all the incoming data
$sanitized = array_map('sanitize', $_POST);
$reason = $sanitized['reason'];
if($reason == "insert"){
$staffs = [];
$stops = [];
$name = $sanitized['rname'];
$code = $sanitized['rcode'];
$desc = $sanitized['rdesc'];
$vnum = $sanitized['vnum'];
$stf = $_POST['staff'];
$st = isset($_POST['stops'])? $_POST['stops']: [];
$st = [];
// foreach($staffs as $staff){
// $stf[] = array_map('sanitize', $staff);
// }
// if(isset($stops)){
// foreach($stops as $stop){
// $st[] = array_map('sanitize', $stop);
// }
// }
$val = insertRoute($conn,$name, $code, $desc, $vnum, $stf, $stops);
echo $val;
}
if($reason == "view"){
$id = $sanitized['id'];
$val = [];
$val = viewRoute($conn,$id);
echo json_encode($val);
}
if($reason == "edit"){
$stf = [];
$stp = [];
$id = $sanitized['pkid'];
$name = $sanitized['rname'];
$code = $sanitized['rcode'];
$desc = $sanitized['rdesc'];
$vnum = $sanitized['vnum'];
$estaffs = $_POST['estaff'];
$estops = $_POST['estops'];
$edel = $_POST['del'];
foreach($estaffs as $val){
$stf[] = array_map('sanitize', $val);
}
foreach($estops as $val){
$stp[] = array_map('sanitize', $val);
}
$cnt = 0;$n_stp = [];
for($i = 0; $i<sizeof($stp); $i++){
if($stp[$i]['stat'] != "Exist"){
$n_stp[$cnt] = $stp[$i];
$cnt++;
}
}
$val = editValues($conn,$id, $name, $code, $desc, $vnum, $stf, $n_stp, $edel);
echo $val;
}
if($reason == "delRoute"){
$id = $sanitized['id'];
$val = delRoute($conn,$id);
echo $val;
}
I tried changing the function to :
function sanitize($input) {
return htmlspecialchars(trim($input));
}
但后来它开始给我以下错误
<b>Warning</b>: trim() expects parameter 1 to be string, array given in <b>C:\xampp\htdocs\gurukul\demo2\controller\routemgmt\route_mgmt.php</b> on line <b>7</b><br />
据我所知,它在修剪中传递数组而不是字符串。 有人可以指导我如何解决这个问题?尝试了一些调试步骤,但没有成功
答案 0 :(得分:0)
在这个函数中,你将在if语句中返回数组,而它应该始终是一个字符串。
<强>替换强>
function sanitize($input) {
if (is_array($input))
return array_map('sanitize', $input);
else
return htmlspecialchars(trim($input));
}
。通过强>
function sanitize($input) {
if (is_array($input))
return sanitize($input);
else
return htmlspecialchars(trim($input));
}
您需要将值返回为字符串而不是数组。不要弄乱$_POST
一个。
全部替换:
array_map('sanitize', $val);
。通过强>
sanitize($val);