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10 interview questions about PHP

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Apr 10, 2018 am 10:43 AM

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This article shares with you the content of 10 interview questions about PHP. Friends in need can refer to it

Q1

The first question is about weak type

$str1 =&#39;yabadabadoo&#39;;
$str2 =&#39;yaba&#39;;
if(strpos($str1,$str2)){
echo"\"". $str1 ."\" contains \"". $str2 ."\"";
}else{
echo"\"". $str1 ."\" does not contain \"". $str2 ."\"";
}

The output result of correct operation:

"yabadabadoo" does not contain "yaba"

strpos is the returned string str2 at the position of str1. If not found, # will be returned. ##falseHowever, this time it actually returned 0 and 0 is also treated as false in the if statement, so we need to make a type judgment on false. The correct code is as follows:

$str1 =&#39;yabadabadoo&#39;;
$str2 =&#39;yaba&#39;;
if(strpos($str1,$str2)!==false){
echo"\"". $str1 ."\" contains \"". $str2 ."\"";
}else{
echo"\"". $str1 ."\" does not contain \"". $str2 ."\"";
}

It should be noted that we use

!==, in php and JS = ! is relative to = = More stringent requirements require consistent data types.

What will the following output be?

$x =5;
echo $x;
echo"<br />";
echo $x+++$x++;
echo"<br />";
echo $x;
echo"<br />";
echo $x---$x--;
echo"<br />";
echo $x;

The actual running result is

5
11
##7
1
##5

About

and

$x--This problem is actually very easy to encounter, we just need to remember$x Use the most recent value before incrementing. The priority of the operator,

is significantly higher than

, so it is executed first and then . Regarding the priority of operators, sometimes we can really use parentheses to make our program more intuitive to understand. After all, the code is not only used for execution. Sometimes the readability of the team may also improve efficiency. A sort of. Q3

About variable references;

$a =&#39;1&#39;;
$b =&$a;
$b ="2$b";

Excuse me

and

$ What is the value of b? The part will immediately come to mind $a='1' $b='21'

, take a closer look

$b=&$a, here $b is a reference to variable $a rather than a direct assignment.

Q4

下面是true还是false

var_dump(0123==123);
var_dump(&#39;0123&#39;==123);
var_dump(&#39;0123&#39;===123);
var_dump(0123 == 123);// false,PHP会默认把0123当作8进制来处理，实际转化为10进制就是83，显然这不是相等的。
var_dump(&#39;0123&#39; == 123);// true这里php会非常有趣的将&#39;0123&#39;转换成一个数字而且默认去掉了前面的0也就是123==123
var_dump(&#39;0123&#39; === 123);// false很显然上面的问题已经说过了数字和字符串类型不一致。

Q5

下面的代码有什么问题吗？输出会是什么，怎样修复它

$referenceTable =array();
$referenceTable[&#39;val1&#39;]=array(1,2);
$referenceTable[&#39;val2&#39;]=3;
$referenceTable[&#39;val3&#39;]=array(4,5);
$testArray =array();
$testArray = array_merge($testArray, $referenceTable[&#39;val1&#39;]);
var_dump($testArray);
$testArray = array_merge($testArray, $referenceTable[&#39;val2&#39;]);
var_dump($testArray);
$testArray = array_merge($testArray, $referenceTable[&#39;val3&#39;]);
var_dump($testArray);

实际输出如下：

array(2){[0]=>int(1)[1]=>int(2)}
NULL
NULL

运行的时候你或许还能看到下面的警告

Warning: array_merge():Argument#2isnot an array
Warning: array_merge():Argument#1isnot an array

array_merge需要传入的参数都是数组，如果不是，则会返回null。你可以这样修改

$testArray = array_merge($testArray,(array)$referenceTable['val1']);
var_dump($testArray);
$testArray = array_merge($testArray,(array)$referenceTable['val2']);
var_dump($testArray);
$testArray = array_merge($testArray,(array)$referenceTable['val3']);
var_dump($testArray);

Q6

$x应该是输出什么？

```
$x =trueandfalse;
var_dump($x);
```

部分同学或许会第一时间想到false,实际上这里依旧是强调运算符的优先级，＝会比 and级别高点，因此等同下面的代码

$x =true;
trueandfalse

答案显而易见。

Q7

经过下面的运算 $x的值应该是多少？

```
$x =3+"15%"+"$25"
```

答案是18，PHP是会根据上下文实现类型的自动转换

上面的代码我们可以这样理解，如果我们在与字符串进行数学运算，实际php会尽可能将字符串中的数组进行转换，如果是数字开头的话则转换成改数字比如"15%"会变成15,如果不是数字开头则会变成0; 上面的运算类似下面：

$x =3+15+0

Q8

运行下面的代码，$text 的值是多少？strlen($text)又会返回什么结果？

$text =&#39;John &#39;;
$text[10]=&#39;Doe&#39;;

上面代码执行完毕后 $text = "John D"(John后面会有连续的5个空格) strlen($text)会返回11

$text[10] = "Doe"给某个字符串具体的某个位置具体字符时候，实际只会把D赋给$text. 虽然$text才开始只有5个自负长度，但是php会默认填充空格。这和别的语言有些差别。

Q9

下面的输出结果会是什么

$v =1;
$m =2;
$l =3;
if( $l > $m > $v){
echo"yes";
}else{
echo"no";
}

实际的输出是"no",只要仔细分析就不难得出

$l>$m 会转换成1 ，则这个时候再和$m比较。

Q10

执行下面代码$x会变成什么值呢？

$x = NULL;
if(&#39;0xFF&#39;==255){
$x =(int)&#39;0xFF&#39;;
}

实际的运行结果是

$x=0

而不是255.

首先'oxFF' == 255我们好判断，会进行转换将16进制数字转换成10进制数字，0xff -> 255.

PHP使用is_numeric_string 判断字符串是否包含十六进制数字然后进行转换。

但是$x = (int)'0xFF';是否也会变成255呢？显然不是，将一个字符串进行强制类型转换实际上用的是convert_to_long,它实际上是将字符串从左向右进行转换，遇到非数字字符则停止。因此0xFF到x就停止了。所以$x=0

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