SQL仅编辑第一个选项

Question

这是我们处理页面中的代码，我们试图在phpmyadmin中编辑数据库，但是即使单击其他条目的“编辑”按钮，似乎也只能编辑我们的第一个条目

require 'dbfunction.php';
$con = getDbConnect();
$crew_id = $_POST["crew_id"];
$CrewName = $_POST["CrewName"];
$CrewRank = $_POST["CrewRank"];
$StartDate = $_POST["StartDate"];
$EndDate = $_POST["EndDate"];
$PayrollNo = $_POST["PayrollNo"];
$EmployeeNo = $_POST["EmployeeNo"];
$WatchKeeping = $_POST["WatchKeeping"];
$Active = $_POST["Active"];
//$Delete = $_POST["Delete"];

if (!mysqli_connect_errno($con)) {

$queryStr = "SELECT crew_id " .
        "FROM crewlist";
}
$result = mysqli_query($con, $queryStr);
while ($row = mysqli_fetch_array($result)) {
if (!mysqli_connect_errno($con)) {
    $sqlQueryStr = "UPDATE crewlist SET crew_name = '$CrewName', crew_rank = '$CrewRank', start_date = '$StartDate' "
            . ", end_date = '$EndDate', payroll_no = '$PayrollNo'"
            . ", employee_no = '$EmployeeNo', watchkeeping = '$WatchKeeping', active = '$Active' WHERE crew_id = " . $row['crew_id'];

    mysqli_query($con, $sqlQueryStr);

    mysqli_close($con);
} else {
    break;
}
header('Location: crewlisting.php');
}

?>

此代码仅编辑数据库的第一项，其余部分保持停滞。 这是我们编辑页面中的代码

<?php
            if (!mysqli_connect_errno($con)) {

                $queryStr = "SELECT * " .
                        "FROM crewlist";
            }
            $result = mysqli_query($con, $queryStr);
            while ($row = mysqli_fetch_array($result)) {

                echo "<tr>.<th>" . $row["crew_name"] . "<br></br>" . "</th>";
                echo "<th>" . $row["crew_rank"] . "</th>";
                echo "<th>" . $row["start_date"] . "</th>";
                echo "<th>" . $row["end_date"] . "</th>";
                echo "<th>" . $row["watchkeeping"] . "</th>";
                echo "<th>" . $row["active"] . "</th>";
                //echo "<td><form action=\"editcrew.php\" method=\"post\"><a href='editcrew.php?del={$row['crew_id']}'>edit</a></form></td>";
                echo "<td><a href=\"editcrew.php?id=" . $row['crew_id'] . "\">Edit</a>";
               echo "<td><form action=\"delete.php\" method=\"post\"><a href='crewlisting.php?del={$row['crew_id']}'>delete</a></form></td>";
            }
            ?>

Answer 1

您将在循环的第一次迭代中关闭mysqli连接。 移动mysqli_close($con); 在循环后面。

编辑：并且您还重定向到循环中的crewlist.php。 如果您重新格式化代码缩进，这将变得显而易见。

Answer 2

从循环中删除mysqli_close($con) ：

while ($row = mysqli_fetch_array($result)) {
    $sqlQueryStr = "UPDATE crewlist SET crew_name = '$CrewName', crew_rank = '$CrewRank', start_date = '$StartDate' "
            . ", end_date = '$EndDate', payroll_no = '$PayrollNo'"
            . ", employee_no = '$EmployeeNo', watchkeeping = '$WatchKeeping', active = '$Active' WHERE crew_id = " . $row['crew_id'];

    mysqli_query($con, $sqlQueryStr);

}
header('Location: crewlisting.php');