我试图将此响应转换为PHP数组：
string(734) " {"definitions":[{"text":"Informal One who is proficient at using or programming a computer; a computer buff.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"Informal One who uses programming skills to gain illegal access to a computer network or file.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"Informal One who enthusiastically pursues a game or sport: a weekend tennis hacker. ","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"See hackie.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"}]}"
我已尝试过
的所有内容 $string = $result;
$pattern = '\w*\(\d+\)\s';
$replacement = '';
preg_replace($pattern,$replacement,$string)
到
$string = $result;
$pattern = '\w*\(\d+\)\s';
$replacement = '';
$def_array = str_replace($pattern,$replacement,$string);
但是当我这样做时，他们甚至不会改变原来的结果。
我希望实现这样的目标：
$def_array = {"definitions":[{"text":"Informal One who is proficient at using or programming a computer; a computer buff.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"Informal One who uses programming skills to gain illegal access to a computer network or file.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"Informal One who enthusiastically pursues a game or sport: a weekend tennis hacker. ","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"},{"text":"See hackie.","attribution":"from The American Heritage\u00ae Dictionary of the English Language, 4th Edition"}]}
我想通过这样的定义：
for($i = 0; $i < sizeof($def_array[0]["definitions"]); $i++)
{
echo $def_array[0]["definitions"][$i]["text"];
echo "\n";
}
我没有使用json_decode，因为当我这样做时，我得到错误Fatal error: Cannot use object of type stdClass as array on the line（$ i = 0; $ i＆lt; sizeof（$ def_array [0] [＆＃34;定义＆＃34} ;]）; $ i ++）`。
答案 0 :(得分：4)
为什么不只是json_decode()函数。
$array = json_decode($input);
在此处查找更多信息：http://www.php.net/manual/en/function.json-decode.php
答案 1 :(得分：0)
Parse error on line 1:
"{ "definitions"
^
Expecting '{', '['
格式无效，必须以{或[
开头