我的表名是tinfo，列名是classconducted。column-conducted类包含这样的内容。
{"segment":["Class I-V Tuition","Class VI-VIII Tuition","Class IX-X Tuition"],"Board":["cbse","cse/Ise","State"],"classFiveSubject":["allsubject","science"],"classeightboard":["cbse","cse/Ise"],"classeightsubject":null,"classTenthboard":["cbse","cse/Ise"],"classTenthsubject":null,"engineering":null}
我想要的输出应该是这样的：
classconducted.
segment:Class I-V Tuition
Board:CBSE,STATE
classFiveSubject:AllSubject,Maths,Science
segment:Class VI-VIII Tuition
Board:CBSE, STATE
classEightSUbject:AllSubject,Maths,Science .
And my index having engineering is null here I don't want to show the null values, can any one guide me. what I have write in query.
<?php
$sql=mysql_query("select * from tinfo");
while($row=mysql_fetch_array($sql))
{
echo $row['classconducted'];
}
?>
有一个名为json_decode()的PHP函数，请参阅手册
基本上，如果你通过这个函数运行json字符串，它会将其转换为PHP变量，即如果json字符串是一个对象，你会得到一个PHP对象，如果它是json数组，你会获得一个PHP数组。。。etc
因此，取您的json字符串并将其通过json_decode（），您将获得
$in = '{"segment":["Class I-V Tuition","Class VI-VIII Tuition","Class IX-X Tuition"],"Board":["cbse","cse/Ise","State"],"classFiveSubject":["allsubject","science"],"classeightboard":["cbse","cse/Ise"],"classeightsubject":null,"classTenthboard":["cbse","cse/Ise"],"classTenthsubject":null,"engineering":null}';
print_r( json_decode($in) );
输出=
stdClass Object
(
[segment] => Array
(
[0] => Class I-V Tuition
[1] => Class VI-VIII Tuition
[2] => Class IX-X Tuition
)
[Board] => Array
(
[0] => cbse
[1] => cse/Ise
[2] => State
)
[classFiveSubject] => Array
(
[0] => allsubject
[1] => science
)
[classeightboard] => Array
(
[0] => cbse
[1] => cse/Ise
)
[classeightsubject] =>
[classTenthboard] => Array
(
[0] => cbse
[1] => cse/Ise
)
[classTenthsubject] =>
[engineering] =>
)
现在，您应该能够使用标准的PHP代码来处理它。