Question

我试图生成一堆同名的输入，并将其保存到sql表中。我遇到的问题是，我的代码适用于大约4个输入，其余的都没有发生。

生成输入的代码：

for ($x = 0; $x < 10; $x++) {
            echo "      ID: <input name=\"IDadd[]\" type=\"text\"> \n"; 
            echo "      Name: <input name=\"Nameadd[]\" type=\"text\"> \n"; 
            echo "      Job: <select name='Jobadd[]'>\n"; 
            echo "      <option value='Tech'>Tech</option>\n";
            echo "      <option value='Check-In'>Check-In</option>\n";
            echo "      <option value='Inventory'>Inventory</option>\n"; 
            echo "      <option value='Beats'>Beats</option>\n"; 
            echo "      <option value='Computers'>Computers</option>\n";
            echo "      <option value='IOS'>IOS</option>\n";
            echo "      <option value='Remarketing'>Remarketing</option>\n";   
            echo "      </select> \n";
            echo "<br><br>";
        }

应将其整理并保存的代码：

if ( isset( $_POST[ 'newEmployee' ] ) )
                            {
                            $index = 0;
                                foreach ($_POST as $key => $value){
                                   $userId = $_POST['IDadd'][$index];
                                   $userName = $_POST['Nameadd'][$index];
                                   if (empty($_POST['Nameadd'][$index])) {
                                    return;
                                  }
                                   $userJob = $_POST['Jobadd'][$index];
                                   $index++;
                                   $sql = "INSERT INTO Employee (userId, Name, Job, Week)
                                    VALUES ('$userId', '$userName', '$userJob', '$currentWeekIs')";
                                    mysqli_query($con, $sql);
                                }
                                if($sql) // will return true if succefull else it will return false
                                    {
                                            mysqli_close( $con );
                                            sleep(4);
                                            echo '<META HTTP-EQUIV="Refresh" Content="0; URL=add.php">';
                                    }
                            }

- 编辑

那么这应该适用吗？

if ( isset( $_POST[ 'newEmployee' ] ) )
                            {

                                if (!empty($_POST['IDadd'])) {
                                    // loop through the keys of one of the arrays
                                    foreach (array_keys($_POST['IDadd']) as $index) {
                                        $userId = mysqli_real_escape_string($_POST['IDadd'][$index]);
                                        $userName = mysqli_real_escape_string($_POST['Nameadd'][$index]);
                                        $userJob = mysqli_real_escape_string($_POST['Jobadd'][$index]);
                                        $sql = "INSERT INTO Employee (userId, Name, Job, Week)
                                        VALUES ('$userId', '$userName', '$userJob', '$currentWeekIs')";
                                        mysqli_query($con, $sql);
                                    }
                                    if($sql) // will return true if succefull else it will return false
                                    {
                                            mysqli_close( $con );
                                            sleep(4);
                                            echo '<META HTTP-EQUIV="Refresh" Content="0; URL=add.php">';
                                    }
                                }
                            }

Answer 1

您的问题是foreach正在遍历$_POST['IDadd']，$_POST['Nameadd']和$_POST['Jobadd']，而不是这些数组的内容。只需查看$key。

的内容

您需要做的是确保它们不是空的，然后遍历内容：

// I'm just showing checking all the arrays aren't empty
// you'll also want to check if they have the same length
if (!empty($_POST['IDadd']) && ...) {
    // loop through the keys of one of the arrays
    foreach (array_keys($_POST['IDadd']) as $index) {
        $userId = mysqli_real_escape_string($_POST['IDadd'][$index]);
        // ...
    }
}

您还应该考虑使用预准备语句来减少SQL注入的可能性。

使用PHP＆amp;生成输入字段SQL

1 个答案: