Question

您好我正在尝试在mysql中插入多个html表行时收到未定义的偏移量：2通知。这是我的html表单代码 `

           <form action="insert_array.php" method="post">                
            <table class="table table-bordered">
                <thead>
                <tr>
                    <th>Name of the degree/certificates</th>
                    <th>Name of the Institution</th>
                    <th>Group</th>
                    <th>Board</th>
                    <th>Year</th>
                    <th>Grade/GPA</th>
                </tr>
                </thead>

                <tr>
                    <td>SSC</td>
                    <td><input name="institution[]" type="text" class="form-control"  ></td>
                    <td><input name="subjectGroup[]" type="text" class="form-control"   ></td>
                    <td><input name="board[]" type="text" class="form-control" ></td>
                    <td><input name="year[]" type="text" class="form-control" ></td>
                    <td><input name="grade[]" type="text" class="form-control"  ></td>

                </tr>

                <tr>
                    <td>HSC</td>
                    <td><input name="institution[]" type="text" class="form-control"  ></td>
                    <td><input name="subjectGroup[]" type="text" class="form-control"  ></td>
                    <td><input name="board[]"  type="text" class="form-control" ></td>
                    <td><input name="year[]" type="text" class="form-control" ></td>
                    <td><input name="grade[]" type="text" class="form-control"  ></td>
                </tr>


            </table>
            <input type="submit" name="submit" value="submit">
        </form>

` 这是我的PHP代码

if($conn === false){
    die("Error: Could not connect. ".mysqli_connect_error());
}
if(isset($_POST['submit'])){
    $i = 0;
    foreach ($_POST as $val){

        $institution = $_POST['institution'][$i];
        $board = $_POST['board'][$i];
        $year = $_POST['year'][$i];
        $subjectGroup = $_POST['subjectGroup'][$i];
        $grades = $_POST['grade'][$i];

        mysqli_query($conn,"INSERT INTO ems_student_qualification (institution,board,year,subjectGroup,grade) VALUES ('$institution', '$board','$year','$subjectGroup','$grades')");
        $i++;
    }
}

＆GT？; 我无法在我的数据库中存储值。但是，当我只用2个字段尝试这个过程时，它工作正常。请指出我的错误是什么？并提前感谢。

Answer 1

您正在循环$_POST而不是多少项

if(isset($_POST['submit'])){
    foreach ($_POST['institution'] as $i => $val){

        $institution = $_POST['institution'][$i];
        $board = $_POST['board'][$i];
        $year = $_POST['year'][$i];
        $subjectGroup = $_POST['subjectGroup'][$i];
        $grades = $_POST['grade'][$i];

        mysqli_query($conn,"INSERT INTO ems_student_qualification (institution,board,year,subjectGroup,grade) VALUES ('$institution', '$board','$year','$subjectGroup','$grades')");

    }
}

Answer 2

尝试使用PHP代码：

if($conn === false){
    die("Error: Could not connect. ".mysqli_connect_error());
}
if(isset($_POST['submit'])){
    $sValues = 'INSERT INTO ems_student_qualification (´institution´, ´board´, ´year´, ´subjectGroup´, ´grade´) VALUES ';
    $i = 0;

    for ($count = count($_POST['institution']); $i < $count; $i++){
        if ( $i > 1 ) { $sValues .= ', '; }

        $institution = mysqli_real_escape_string($conn, $_POST['institution'][$i]);
        $board = mysqli_real_escape_string($conn, $_POST['board'][$i]);
        $year = mysqli_real_escape_string($conn, $_POST['year'][$i]);
        $subjectGroup = mysqli_real_escape_string($conn, $_POST['subjectGroup'][$i]);
        $grades = mysqli_real_escape_string($conn, $_POST['grade'][$i]);

        $sValues .= '(´' . $institution . '´, ´' . $board . '´, ´' . $year . '´, ´' . $subjectGroup . '´, ´' . $grades . '´)';
    }

    if ( $i > 0 ) {
        mysqli_query($conn, $sValues);
    }
}

它还包括字符串转义。

Answer 3

if(isset($_POST['submit'])){
   for ($i=0; $i < count($_POST['institution']); $i++ )
  {
     $institution = $_POST['institution'][$i];
    $board = $_POST['board'][$i];
    $year = $_POST['year'][$i];
    $subjectGroup = $_POST['subjectGroup'][$i];
    $grades = $_POST['grade'][$i];

    mysqli_query($conn,"INSERT INTO ems_student_qualification (institution,board,year,subjectGroup,grade) VALUES ('$institution', '$board','$year','$subjectGroup','$grades')");
  }}

Answer 4

你的循环不合适，因为它减少了总变量的数量，即11次，但是$ _POST [＆＃39;机构＆＃39;]只存在2次（数组从零开始）所以第3次它显示{{ 1}}。可以使用undefined error或任何输入字段值进行循环，例如：$_POST['institution']，也可以使用for循环。希望这很清楚。

mysql表

4 个答案: