PHP數組不插入多行到數據庫

Question

我已經創建了複雜的表單，我會嘗試提供簡化的示例。通過點擊+按鈕可以生成更多字段。

例如在形式上是字段：

通過

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

點擊+按鈕將其添加重複行（通過JavaScript），因此點擊形式的+按鈕部分後的樣子：

NameOfVessel TypeOfVessel  YearBuilt 
[   ] [   ] [   ] 

NameOfVessel TypeOfVessel  YearBuilt 
[   ] [   ] [   ] + 

有可以按用戶需要多次點擊+按鈕。

我有HTML格式是這樣的：

<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插入到數據庫中。它應該將所有添加的行中的值插入到多個數據庫表的行中，但現在它不插入任何內容。

$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); 

我用var_dump，但它返回NULL。這段代碼有什麼問題？你有什麼想法嗎？

UPDATE

我試着在下面做：

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" /> 

但在這種情況下，我得到了Id = ' + 'NameOfVessel' + noOfClicks + '。據我瞭解，我需要通過JavaScript來進行連接，只是我無法正確實現。

Answer 1

串mysqli_real_escape_string（mysqli的$鏈路，字符串$ escapestr）

（docs）

mysqli_real_escape_string期望一個字符串，而不是陣列。

您應該先循環$_POST['NameOfVessel']數組並應用mysqli_real_escape_string上的值。其他帖子鍵也一樣。

假設$_POST['NameOfVessel']，$_POST['TypeOfVessel']和$_POST['YearBuilt']有相同數量的元素，你可以這樣做：

$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); 
    } 
} 

要克隆後達到標識的唯一性，看到這樣的回答：jQuery clone and change Ids。

它需要一些適應性。也許更容易完全刪除id。