javascript 如果变量等于数字 1
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If variable equals the number 1
提问by loriensleafs
I'm trying to write a very simple if statement in my jquery. If the variable equals the number one then do this, if it equals the number two then do this. This is what I wrote but it doesn't seem to work and I can't figure out why:
我正在尝试在我的 jquery 中编写一个非常简单的 if 语句。如果变量等于数字 1，则执行此操作，如果它等于数字 2，则执行此操作。这是我写的，但它似乎不起作用，我不知道为什么：
$("#next-btn").click(function() {
if (current_slide = 1) {
first_screen();
}, else if (current_slide = 2) {
second_screen();
}
});
probably very simple, but I appreciate the help.
可能很简单，但我很感激你的帮助。
回答by Adil
You need to use comparison operator==in if statementcondition instead of assignment operator =also remove commaafter first closing curly bracket of then(true) block of ifstatement. You can test it over here.
您需要在if 语句条件中使用比较运算符而不是赋值运算符，并且在(true)语句块的第一个关闭大括号后也删除。你可以在这里测试它。===commathenif
if (current_slide == 1) {
first_screen();
} else if (current_slide == 2) {
second_screen();
}
I assume current_slide has some number to compare, you read below how comparison operator == performs the comparison.
我假设 current_slide 有一些要比较的数字，您在下面阅读了比较运算符 == 如何执行比较。
Comparion equal operator
比较等号运算符
If the two operands are not of the same type, JavaScript converts the operands then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the other operand is converted to a string if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory, reference.
如果两个操作数的类型不同，JavaScript 会转换操作数，然后应用严格的比较。如果操作数是数字或布尔值，则尽可能将操作数转换为数字；else 如果任一操作数是字符串，则另一个操作数在可能的情况下转换为字符串。如果两个操作数都是对象，然后JavaScript的比较内部引用时的操作数是指在存储器中的相同对象，这是相等的参考。
回答by Nishu Tayal
Use comparison operator "==" instead of assignment "=" and remove comma before else if.
使用比较运算符“==”而不是赋值“=”并在else if.
$("#next-btn").click(function() {
if (current_slide == 1) {
first_screen();
}else if (current_slide == 2) {
second_screen();
}
});
回答by falinsky
May be it'd be better using switch:
使用 switch 可能会更好：
$("#next-btn").click(function() {
switch(current_slide) {
case 1:
first_screen();
break;
case 2:
second_screen();
break;
}
});
回答by lfergon
Youd shouldn′t check a condition with an assigment:
你不应该用赋值来检查条件：
$("#next-btn").click(function() {
if(current_slide===1){
first_screen();
}else if (current_slide===2){
second_screen();
}
});
