php - 有一个数组到字符串的转换错误

Question

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

下面的代码试图获取 MYSQL 表并为表创建标题和列以打印在 pdf 页面上。

 $result=mysqli_query($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";

}

从上面的代码我得到一个错误提示

Notice: Array to string conversion in C:\xampp\htdocs\Namtax\leave_view.php on line 64

这是哪一行

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

而且我似乎不明白为什么错误仅针对该行显示而不显示其余部分以及有关如何修复它的任何帮助？

Answer 1

你有这个代码

 $compassionate_leave = ["compassionate_leave"];

其数组，改为

$compassionate_leave = $row["compassionate_leave"];