PHP数组未向数据库插入多行

Question

我已经创建了复杂的表格，我将尝试提供简化的示例。通过单击+ 按钮可以生成更多字段。

例如在表单中是字段：

Certificate    Date Of Issue    Date of Expire   
[         ]    [           ]    [            ]   +

通过单击+ 按钮它添加重复行（通过javascript），因此在单击+ 按钮后，表单的一部分看起来像：

NameOfVessel    TypeOfVessel       YearBuilt  
[          ]    [           ]    [            ]

NameOfVessel    TypeOfVessel       YearBuilt  
[          ]    [           ]    [            ]   +

可以根据用户需要多次单击+ 按钮。

我有这样的 HTML 表单：

<li>
    <ul class="column">         
        <li>
            <label for="NameOfVessel">Name of Vessel</label>
            <input id="NameOfVessel" type="text" name="NameOfVessel[]" class="field-style field-split25 align-left" placeholder="Name of Vessel" /> 
        </li>
    </ul>
</li>
<li>
    <ul class="column">         
        <li>
            <label for="TypeOfVessel">Type of Vessel</label>
            <input id="TypeOfVessel" type="text" name="TypeOfVessel[]" class="field-style field-split25 align-left" placeholder="Type of Vessel" /> 
        </li>           
    </ul>
</li>
<li>
    <ul class="column">         
        <li>
            <label for="YearBuilt">Year Built</label>
            <input id="YearBuilt" type="text" name="YearBuilt[]" class="field-style field-split25 align-left" placeholder="Year Built" />   
        </li>           
    </ul>
</li>

PHP 插入数据库。它应该将所有添加的行中的值插入到多个数据库表的行中，但现在它没有插入任何东西。

$UserID = get_current_user_id();
$NameOfVessel = mysqli_real_escape_string($link, $_POST['NameOfVessel']);       
$TypeOfVessel = mysqli_real_escape_string($link, $_POST['TypeOfVessel']);       
$YearBuilt = mysqli_real_escape_string($link, $_POST['YearBuilt']); 

foreach($NameOfVessel as $key=>$res) {
    $sql2 = "INSERT INTO CV_SeaServices (NameOfVessel, UserId, TypeOfVessel, YearBuilt) VALUES ('$res', '$UserId[$key]', '$TypeOfVessel[$key]', '$YearBuilt[$key]')";
    if(mysqli_query($link, $sql2)){
        echo "Resume created successfully.";
    } else {
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }
}
var_dump($NameOfVessel);

我使用了var_dump，但它返回NULL。这段代码有什么问题？你有什么想法吗？

更新

我尝试过以下操作：

JS：

var noOfClicks = 0;
$(document).ready(function() {
    $(".add-row").click(function() {
        $("ul.sea-service").first().clone().appendTo(".personal-details1").append('<button class="remove">X</button>').find('input').val('');
        noOfClicks += 1;

    });
    $("body").on('click', '.remove', function() {
        $(this).closest('.sea-service').remove();
    });
});

HTML：

<input id="NameOfVessel' + noOfClicks + '" type="text" name="NameOfVessel[]" class="field-style field-split25 align-left" placeholder="Name of Vessel" />

但在这种情况下，我得到了 Id = ' + 'NameOfVessel' + noOfClicks + '。据我了解，我需要通过 javascript 进行连接，只是我无法正确实现它。

Answer 1

string mysqli_real_escape_string (mysqli $link, string $escapestr)

(docs)

mysqli_real_escape_string 需要一个字符串，而不是一个数组。

您应该首先循环 $_POST['NameOfVessel'] 数组并将 mysqli_real_escape_string 应用于值。其他帖子键也是如此。

假设$_POST['NameOfVessel']、$_POST['TypeOfVessel'] 和$_POST['YearBuilt'] 具有相同数量的元素，您可以执行以下操作：

$userId = $UserId[$key]; // because you're overriding `$key` below.
foreach($_POST['NameOfVessel'] as $key => $val){
    $NameOfVessel = $val;
    $TypeOfVessel = $_POST['TypeOfVessel'][$key];
    $YearBuilt    = $_POST['YearBuilt'][$key];

    $NameOfVessel = mysqli_real_escape_string($link, $NameOfVessel); 
    $TypeOfVessel = mysqli_real_escape_string($link, $TypeOfVessel); 
    $YearBuilt    = mysqli_real_escape_string($link, $YearBuilt); 

    $sql2 = "INSERT INTO CV_SeaServices 
            (NameOfVessel, UserId, TypeOfVessel, YearBuilt) 
            VALUES 
            ('$res', '$userId', '$TypeOfVessel', '$YearBuilt')";

    if(mysqli_query($link, $sql2)){
        echo "Resume created successfully.";
    } else {
        echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
    }
}

要在克隆后实现 ID 的唯一性，请参阅此答案：jQuery clone and change Ids。

它需要一些适应。也许完全删除 id 会更容易。