Prompt for undefined variables: gradeList. How can you solve this problem? I beg you for answers.
轻微强迫症
轻微强迫症 2018-05-27 16:09:23
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轻微强迫症
轻微强迫症

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kiraseo
//定义可以在前面定义gradeList空数组就可以了
//例如
$gradeList=[];

This will not prompt the error of undefined variables

手机用户067795602

You can define the gradeList variable in front

$gradeList=null It should be because there is no data in your table $grade has no data and no foreach loop is performed

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