Inside the hover event handler, this is going to be an li, since it is attached to $(‘.smenu li’). Then $(this).find(‘li’) will find li elements inside .smenu li, which presumably don’t exist. solved If else is not workin in Jquery? [closed]
Use || statement. <?php if($CURRENT_USER[‘selected_plan_id’] == ‘4’ || $CURRENT_USER[‘selected_plan_id’] == ‘5’ || $CURRENT_USER[‘selected_plan_id’] == ‘6’): ?> Or Use in_array(). $arr = Array (4,5,6); <?php if(in_array($CURRENT_USER[‘selected_plan_id’],$arr)): ?> 1 solved PHP if statement with OR operators [closed]
You have to write if($_POST[‘postnummer’] == “7900” || $_POST[‘postnummer’] == “7950” || $_POST[‘postnummer’] == “7960”) { $region = “Nordjylland”; } elseif ($_POST[‘postnummer’] == “8654” || $_POST[‘postnummer’] == “8660” || $_POST[‘postnummer’] == “8680” || $_POST[‘postnummer’] == “8700”) { $region = “Midtjylland”; } solved php if elseif statement select wrong variable every time [closed]
it should be the following for it to work correctly: var = “1” if var == “1”: print(“DL”) else: print(“DSU”) As said in the comments, print() is a function so to check if the “1” matches the condition, put it in a variable. 0 solved Python – If Statment Not Working Properly [closed]
if you’re only comparing a few values then you might as well proceed with the current approach as there is nothing in place to make it shorter. However, if you’re repeating your self many times, then you can create a helper function to do the work for you. i.e static boolean anyMatch(int comparisonValue, int… elements){ … Read more
It will print out 5. It doesn’t even approach to the second if-else, because doesn’t pass the first condition x!=5 and the else statement is missing and it goes to the last line where you print the variable out. It’s the same as: int x=5, y=5, z=5; if (x!=5) { if (y<=7) { z=z+4; } … Read more
your value at key[18] is “L” so if condition is always true and you will get an alert with value 7 1 solved Javascript for loop with if statement not reaching the else if statement [closed]
Your Question is Simple.. If and else are used, When you make a true or false statements. Example: If ( 3>2) { printf (“3”);} else { printf ( “2”);} But in case of if and else if Example: if(3>2) {printf (“3”);} else if (2>3) {printf (“false”);} IN THIS WAY .. In else if. you should … Read more
I think you need something like this: // Make sure x is smaller than y if(x>y) { swap(&x, &y); } // Make sure x is smaller than z if(x>z) { swap(&x, &z); } // Now x is smaller than both y and z // Make sure y is smaller than z if(y>z) { swap(&y, &z); … Read more
The if condition checking must be done using == or === operator.Because = is used to assign something.So try this if($d == 2){ $dayznum = $dayznum2; } 3 solved PHP if($d=2) not working
You are breaking the loop and your if(valid) code is within the for loop. You probably meant to have the if(valid) outside the for loop scope. For example you can do: valid = false; for (var i = 0; i < usernameArray.length; i++) { if ((un == usernameArray[i]) && (pw == passwordArray[i])) { valid = … Read more
You need to check each case individually using the OR operator if you want to check more than one case. if (BothMods == “Yes” || BothMods == “YES” || BothMods == “yes”) { // do whatever } 4 solved c++ if specific words from string
In your code, if (number == 10 || 11 || 12 || 13 || 14 || 15 || 16 || 17 ||18 || 19) is a wrong approach. You cannot chain the logical operator like that. As of now, because of the operator precedence, your code is essentially if ( (number == 10) || 11 … Read more
Your code is equivalent to: double x1 = 0.9; float xf = (float)x1; // discard some precision double x2 = (double)xf; if (x2 >= x1) { // … } The problem is that, after you discarded some precision from x1 by converting it to float, it can never be recovered; casting it back to double … Read more
The correct syntax is if condition: # stuff elif other: # stuff elif some_other: # stuff else: # stuff Note that else does not get any explicit condition, it is the catch-all if none of the above conditions were True. 3 solved Proper if..else syntax in Python