PHP How to fix Notice Undefined variable

Question

code   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       Error    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       how to fix that

User · Answer

You should initialize your variables outside the while loop. Outside the while loop, they currently have no scope. You are just relying on the good graces of php to let the values carry over outside the loop

           $hn = "";
           $pid = "";
           $datereg = "";
           $prefix = "";
           $fname = "";
           $lname = "";
           $age = "";
           $sex = "";
           while (...){}

alternatively, it looks like you are just expecting a single row back. so you could just say

$row = pg_fetch_array($result);
if(!row) {
    return array();
}
$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) ;

User · Answer

I would guess your query isn t running as expected and you are getting to the return line with undefined variables   Also  the way you are doing the variable assignment  you would be overwriting the same variable with each loop iteration  so you wouldn t return the entire result set   Finally  it seems odd to return a numerically-keyed result set instead of an associatively-keyed one   Consider naming only the fields needed in the SELECT and keeping the key assignments   So something like this   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          return   array       while   row   pg fetch array  result            return      row                     return  return      You  might also consider opening a question about how to improve your query  is it is pretty heinous as it stands now

User · Answer

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

User · Answer

It looks like you don t have any records that match your query  so you d want to return an empty array  or null or something  if the number of rows    0

User · Answer

Xampp I guess you re using MySQL  mysql fetch array  result      And make sure  result is not empty

User · Answer

Declare them before the while loop    hn        pid        datereg        prefix        fname        lname        age        sex         You are getting the notice because the variables are declared and assigned inside the loop

[php] PHP How to fix Notice: Undefined variable:

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