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Small problems caused by array foreach

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2016-07-29 09:09:41882browse

Code

<code><span>$arr1</span> = [ <span>1</span>, <span>2</span>, <span>3</span>, <span>4</span>, <span>5</span> ];
<span>$arr2</span> = [ <span>'a'</span>, <span>'b'</span>, <span>'c'</span>, <span>'d'</span>, <span>'e'</span> ];
<span>$arr3</span> = [];
<span>foreach</span> (<span>$arr1</span><span>as</span> & <span>$v</span>){
    <span>$v</span> += <span>10</span>; 

}

<span>foreach</span> (<span>$arr2</span><span>as</span><span>$k</span> => <span>$v</span>){
    <span>//举例</span><span>$v</span> = <span>$v</span> . <span>$arr1</span>[ <span>$k</span> ];
    <span>$arr3</span>[ <span>$k</span> ] = <span>$v</span>;
}
<span>echo</span> implode(<span>', '</span>, <span>$arr1</span>) . <span>"\n"</span> . implode(<span>', '</span>, <span>$arr2</span>) . <span>"\n"</span> . implode(<span>', '</span>, <span>$arr3</span>);</code>

Run

<code><span>11</span>, <span>12</span>, <span>13</span>, <span>14</span>, ee
<span>a</span>, b, c, d, e
a11, b12, c13, d14, ee</code>

Result

The reason for the problem is that after the end of the first loop, the corresponding $v was not released

Solve

before the loop, through unset() , release the variables and this problem will not occur

Between the two loops, add unset($v);

<code><span>11</span>, <span>12</span>, <span>13</span>, <span>14</span>, <span>15</span><span>a</span>, b, c, d, e
a11, b12, c13, d14, e15</code>
').addClass('pre-numbering').hide() ; $(this).addClass('has-numbering').parent().append($numbering); for (i = 1; i ').text(i)); }; $numbering.fadeIn(1700); }); });

The above introduces the minor problems caused by array foreach, including the content. I hope it will be helpful to friends who are interested in PHP tutorials.

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