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php弱类型比较

Jun 06, 2016 pm 08:14 PM
php

var_dump(0) //int(0)
var_dump('0') //string(1) "0"
var_dump('ssddsd') //string(7) "sasadad"
var_dump(0=='sdfsdf') // bool(true)
var_dump('0'=='sdfsf') //bool(false)

在上面这些比较中为什么var_dump(0=='sdfsdf') // bool(true)而var_dump('0'=='sdfsf') //bool(false)
我认为的是0是整形,和字符型'sdfsf'比较应该是false。。。。

为什么实际结果却是相反的,我哪里理解错了么???

回复内容:

var_dump(0) //int(0)
var_dump('0') //string(1) "0"
var_dump('ssddsd') //string(7) "sasadad"
var_dump(0=='sdfsdf') // bool(true)
var_dump('0'=='sdfsf') //bool(false)

在上面这些比较中为什么var_dump(0=='sdfsdf') // bool(true)而var_dump('0'=='sdfsf') //bool(false)
我认为的是0是整形,和字符型'sdfsf'比较应该是false。。。。

为什么实际结果却是相反的,我哪里理解错了么???

<code>var_dump(0=='sdfsdf')</code>
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当处理这句时,不同类型比较'sdfsdf'会被强制转换成int型,也就是int 0,所以结果是true

<code>var_dump('0'=='sdfsf') </code>
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当处理这句时,相同类型比较直接比较内容,内容不一样,所以结果是false

你可以试试var_dump((int)("sdfsf"));结果是int 0

https://segmentfault.com/q/1010000000095573/a-1020000000098176

至于后面那个为啥错误的… 这个还需要说明么… 判断相等首要的因素是内容要相等,只是 == 非严格模式会帮我们自动将两者的类型转换成一致的而已。字符串内容都不同你到底是为啥会觉得相等的…

你理解错了。
比较运算符直接的转换 是:整数和字符串比较,字符串转为整数,那么字符串sdfsdf就转为整数0了,所以第三条是true。
第四条是true因为他们都是字符串,所以就直接比较了,那么字符串0和sdfsf自然不等,为false

==比较只比较值,不同类型会转换为同一类型比较。

要比较类型用===,必须值和类型都一样才为true。

这些官方文档里都有写的,请仔细阅读文档。

var_dump(0=='sdfsdf'); 弱类型比较通常会先进行类型转换转为相同类型在进行值相等判断.
其中字符串和数值的比较规则是字符串转换成数字. 所以上面那个比较实际的进行可看作

var_dump(0 == (int)'sdfsdf');
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var_dump('0'=='sdfsf');两个都是字符串就是比较值是否相等.

字符串和数字关键后会被转换成0,这样的现象在数据库中也会存在的

http://php.net/manual/zh/language.operators.comparison.php
如果比较一个数字和字符串或者比较涉及到数字内容的字符串,
则字符串会被转换为数值并且比较按照数值来进行.此规则也适用于switch语句.
当用===或!==进行比较时则不进行类型转换,因为此时类型和数值都要比对.

PHP中字符串跟数字比较时,字符串会被转化为数字后进行比较,转化过程类似intval():

<code>echo intval('5stars');  输出5
print_r('5stars'==5);   输出1
var_dump('5stars'==5);  输出bool(true)
var_dump('5stars'===5); 输出bool(false)
var_dump('0e1'=='0e2');  输出bool(true)
var_dump('0e1'==='0e2'); 输出bool(false)
</code>
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PHP中比较字符串应该用恒等于===或者strcmp.

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