如何将 js 表格单元格值发布到 PHP 并插入到 MYSQL 数据库

Question

我有一个动态 HTML 表，我只是使用 JavaScript 调用 id，遍历数据并捕获单元格值。

问题- 我如何将此数据发布到测试 save.php 页面以将整个表插入我的 MySQl 数据库？

    var table = document.getElementById('table'), 
    rows = table.getElementsByTagName('tr'),
    i, j, cells, OS_d,roleApp_d,Men_d,vcpu_d,Val_d,Per_d,hw_d ;

    for (i = 1, j = rows.length; i < j; ++i) {
    cells = rows[i].getElementsByTagName('td');
    if (!cells.length) {
    continue;
    }

    OS_d = cells[1].firstChild.value;
    roleApp_d = cells[2].firstChild.value;
    vcpu_d = cells[3].firstChild.value;
    Men_d = cells[4].firstChild.value;
    Val_d = cells[5].firstChild.value;
    Per_d = cells[6].firstChild.value;
    hw_d = cells[7].firstChild.value;

    }

--- 需要帮助的部分

    $.ajax({
    type: "POST",
    url: 'Save.php',
    data: ({data:OS_d, roleApp_d, vcpu_d, Men_d, Val_d, Per_d, hw_d}),
    success: function(data) {
    alert(data);
    }
    });

PHP 端保存.php

    <?php
    $test = $_POST['data'];
    echo $test;
    ?>

Answer 1

Just quick code, not tested it so your post data should be like this:
    var table = document.getElementById('table'),
    rows = table.getElementsByTagName('tr'),
    i, j, cells, OS_d,roleApp_d,Men_d,vcpu_d,Val_d,Per_d,hw_d ;

    var p_row = '[';
    for (i = 1, j = rows.length; i < j; ++i) {
       cells = rows[i].getElementsByTagName('td');
       if (!cells.length) {
          continue;
       }

        p_row += "{";
        p_row +='"OS_d":"'+cells[1].firstChild.value+'",';
        p_row +='"roleApp_d":"'+ cells[2].firstChild.value+'",';
        p_row +='"vcpu_d":"'+ cells[3].firstChild.value+'",';
        p_row +='"Men_d":"'+ cells[4].firstChild.value+'",';
        p_row +='"Val_d":"'+ cells[5].firstChild.value+'",';
        p_row +='"Per_d":"'+ cells[6].firstChild.value+'",';
        p_row +='"hw_d":"'+ cells[7].firstChild.value+'"';
        p_row += '},';

    }
    p_row = (p_row).replace(new RegExp("[,]+$"), "");
    p_row += ']';

   var formData = new FormData();
   formData.append('data',p_row);
   fetch("/save.php", {
            method: "POST",
            body: formData
        }).then(function (response) {
            console.log(response);

        }).catch(function (error) {
            console.error(error, "code: " + error.status);
   }); 

Answer 2

我想通了，所以我决定与社区分享。 感谢您的支持。

   $.ajax({
    type: "POST",
    url: 'test.php',
    data: {p_row  : p_row },
    success: function(data){
        alert(data);
    }
    });
    }

    On the test.php side

    <?php

    $data = false;
    if(isset($_POST['p_row'])){
        $data  = $_POST['p_row'];
    }

    echo 'Received Data: ' . $data ;
    //$conn->close();
    ?>