[英]PHP, MYSQL workbench, postman not working correctly
I have 2 users in a table, for example Vesna and Silvija, Vesna is first, Silvija second and if I type Silvija in postman he will return me "User does not exist and than Id of a user Silvija. This should be working on a way that Postman return just an Id of a User and if I type user that doesnt exist he should return that user doesnt exist. But when I type a users that doesnt actually exist he will return 2 times that users doesnt exist instead one time because table have 2 users that doesnt match entered user. Hope I have explained it understandably and that someone can help me. 我在一个表中有2位用户,例如Vesna和Silvija,Vesna是第一位,Silvija第二位,如果我在邮递员中键入Silvija,他将返回我“用户不存在,并且比用户Silvija的ID大。这应该在邮递员只返回一个用户ID的方式,如果我键入一个不存在的用户,他应该返回该用户不存在,但是当我键入一个不存在的用户时,他将返回该用户不存在的2次而不是一次,因为表有2位用户与输入的用户不匹配。希望我已经理解了,并且有人可以帮助我。
foreach($oUsers as $oUser)
{
if($oUser->USERNAME == $sUsername)
{
$_SESSION['user_id'] = $oUser->USER_ID;
$_SESSION['username'] = $oUser->USERNAME;
echo $_SESSION['user_id'];
break;
}
else
{
echo ' User does not exist';
}
}
You can not use echo '...'
inside the foreach loop because echo
is executed in every iteration. 您不能在foreach循环中使用
echo '...'
,因为echo
在每次迭代中都执行。
$found = false;
foreach($oUsers as $oUser){
if($oUser->USERNAME == $sUsername){
$_SESSION['user_id'] = $oUser->USER_ID;
$_SESSION['username'] = $oUser->USERNAME;
$found = true;
break;
}
}
if($found === true){
echo $_SESSION['user_id'];
}else{
echo ' User does not exist';
}
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