Php 无法从类函数调用中获取返回的布尔值
我无法从php函数调用中获取返回值Php 无法从类函数调用中获取返回的布尔值,php,Php,我无法从php函数调用中获取返回值 ... $dey=$validate->yoxla($this->ad,$this->soyad,$this->istadi,$this->sifre,$this->email); if($dey){ // i cant get returned value from here. even it is true. ech
...
$dey=$validate->yoxla($this->ad,$this->soyad,$this->istadi,$this->sifre,$this->email);
if($dey){ // i cant get returned value from here. even it is true.
echo 'asd';
if(!$checkuser->checkusername($this->istadi)){
$err=array();
$err['code']='code14';
$err['aciglama']=$code14;
print json_encode($err);
die();
}elseif(!$checkuser->checkemail($this->email)){
$err=array();
$err['code']='code15';
$err['aciglama']=$code15;
print json_encode($err);
die();
}else{
$err=array();
$err['code']='acig';
print json_encode($err);
die();
}
}
yoxla()函数:
...
elseif (!preg_match("/^.*(?=.{8,})(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z]).*$/", $sifre)){
$err=array();
$err['code']='code11';
$err['aciglama']=$code11;
print json_encode($err);
die();
}elseif($email==""){
$err=array();
$err['code']='code12';
$err['aciglama']=$code12;
print json_encode($err);
die();
}elseif (!$this->validEmail($email)) {
$err=array();
$err['code']='code13';
$err['aciglama']=$code13;
print json_encode($err);
die();
}else{
return true;
}
我通过ajax请求调用第一个。并且没有任何东西返回到ajax,或者我无法从getresult()函数中获取返回值 从类声明中删除():
试试这个:
class Form{
public function getresult(){
return true;
}
}
$validate=new Form();
$result=$validate->getresult();
if($result){
echo 'true';
}else{
echo 'false';
}
启用php错误日志记录,以便您能够在浏览器上理解此类错误:您应该第一次发布正确的代码,现在答案与问题不匹配。对不起,这是我的错(您可以创建一个新问题,因为现在此答案无效):\