einen Array String Konvertierungsfehler mit

Question

$conn = mysqli_connect("$host", "$username", "$password")or die("cannot connect"); 
mysqli_select_db($conn,"$db_name")or die("cannot select DB"); 

der folgenden Code versucht, die MYSQL-Tabelle zu erhalten und die Header und Spalten für die Tabelle auf einer PDF-Seite zu drucken, erstellen.

$result=mysqli_query($conn,"select Employee_number,date_start,date_end,Days_taken,Sick,Study,Annual,compassionate_leave,Other,Details,Status,approved_by from $tbl_name "); 

$number_of_products = mysqli_num_rows($result); 

//Initialize the 3 columns and the total 
$column_Employee_number = ""; 
$column_date_start = ""; 
$column_date_end = ""; 
$column_Days_taken = ""; 
$column_Sick = ""; 
$column_Study = ""; 
$column_Annual = ""; 
$column_compassionate_leave = ""; 
$column_Other = ""; 
$column_Details = ""; 
$column_Status = ""; 
$column_approved_by = ""; 

$total = 0; 

//For each row, add the field to the corresponding column 
while($row = mysqli_fetch_array($result)) 
{ 
    $Employee_number = $row["Employee_number"]; 
    $date_start = $row["date_start"]; 
    $date_end = $row["date_end"]; 
    $Days_taken = $row["Days_taken"]; 
    $Sick = $row["Sick"]; 
    $Study = $row["Study"]; 
    $Annual = $row["Annual"]; 
    $compassionate_leave = ["compassionate_leave"]; 
    $Other = $row["Other"]; 
    $Details = $row["Details"]; 
    $Status = $row["Status"]; 
    $Other = $row["Other"]; 
    $approved_by =$row["approved_by"]; 


    $column_Employee_number =$column_Employee_number.$Employee_number."\n"; 
    $column_date_start = $column_date_start.$date_start."\n"; 
    $column_date_end = $column_date_end.$date_end."\n"; 
    $column_Days_taken = $column_Days_taken.$Days_taken."\n"; 
$column_Sick = $column_Sick.$Sick."\n"; 
$column_Study = $column_Study.$Study."\n"; 
$column_Annual = $column_Annual.$Annual."\n"; 
$column_compassionate_leave = $column_compassionate_leave.$compassionate_leave."\n"; 
$column_Other = $column_Other.$Other."\n"; 
$column_Details = $column_Details.$Details."\n"; 
$column_Status = $column_Status.$Status."\n"; 
$column_approved_by = $column_approved_by.$approved_by."\n"; 

} 

aus dem obigen Code bekomme ich einen Fehler

Hinweis sagen: Array String-Konvertierung in C: \ xampp \ htdocs \ Namtax \ leave_view.php on line 64

Das ist diese Zeile

$column_compassionate_leave = $column_compassionate_leave.$compassionate_leave."\n"; 

und ich nicht zu verstehen, warum der Fehler angezeigt wird ly für diese Linie und nicht für den Rest und irgendeine Hilfe, wie man es repariert?

Answer 1

Sie haben diesen Code

$compassionate_leave = ["compassionate_leave"]; 

seine Anordnung, Änderung

$compassionate_leave = $row["compassionate_leave"]; 

Answer 2

Änderung Ihrer $compassionate_leave = ["compassionate_leave"]; sollte es $compassionate_leave = $row["compassionate_leave"];

Answer 3

$compassionate_leave = ["compassionate_leave"]; 

Diese Linie

wird sein