PHP如何修复注意:未定义的变量: [英] PHP How to fix Notice: Undefined variable:

问题描述

代码:

Function ShowDataPatient($idURL)
{
    $query =" select * from cmu_list_insurance,cmu_home,cmu_patient where cmu_home.home_id = (select home_id from cmu_patient where patient_hn like '%$idURL%')
                     AND cmu_patient.patient_hn like '%$idURL%'
                     AND cmu_list_insurance.patient_id like (select patient_id from cmu_patient where patient_hn like '%$idURL%') ";

    $result = pg_query($query) or die('Query failed: ' . pg_last_error());

    while ($row = pg_fetch_array($result))
    {
        $hn = $row["patient_hn"];
        $pid = $row["patient_id"];
        $datereg = $row["patient_date_register"];
        $prefix = $row["patient_prefix"];
        $fname = $row["patient_fname"];
        $lname = $row["patient_lname"];
        $age = $row["patient_age"];
        $sex = $row["patient_sex"];
    }          
    return array($hn,$pid,$datereg,$prefix,$fname,$lname,$age,$sex);
}

错误:

Notice: Undefined variable: hn in C:\xampp\htdocs\...  
Notice: Undefined variable: pid in C:\xampp\htdocs\... 
Notice: Undefined variable: datereg in C:\xampp\htdocs\...    
Notice: Undefined variable: prefix in C:\xampp\htdocs\...    
Notice: Undefined variable: fname in C:\xampp\htdocs\...    
Notice: Undefined variable: lname in C:\xampp\htdocs\...    
Notice: Undefined variable: age in C:\xampp\htdocs\...    
Notice: Undefined variable: sex in C:\xampp\htdocs\...

该如何解决?

推荐答案

在函数的开头定义变量，因此，如果没有记录，则该变量存在，并且不会出现错误.在返回的数组中检查空值.

Define the variables at the beginning of the function so if there are no records, the variables exist and you won't get the error. Check for null values in the returned array.

$hn = null;
$pid = null;
$datereg = null;
$prefix = null;
$fname = null;
$lname = null;
$age = null;
$sex = null;

这篇关于PHP如何修复注意:未定义的变量:的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！