我已经创建了复杂的表单,我会尝试提供简化的示例。通过点击+
按钮可以生成更多字段。PHP数组不插入多行到数据库
例如在形式上是字段:
通过Certificate Date Of Issue Date of Expire
[ ] [ ] [ ] +
点击+
按钮将其添加重复行(通过JavaScript),因此点击形式的+
按钮部分后的样子:
NameOfVessel TypeOfVessel YearBuilt
[ ] [ ] [ ]
NameOfVessel TypeOfVessel YearBuilt
[ ] [ ] [ ] +
有可以按用户需要多次点击+
按钮。
我有HTML格式是这样的:
<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插入到数据库中。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,但现在它不插入任何内容。
$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
。这段代码有什么问题?你有什么想法吗?
UPDATE
我试着在下面做:
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() {
$(this).closest('.sea-service').remove();
});
});
HTML:
<input id="NameOfVessel' + noOfClicks + '" type="text" name="NameOfVessel[]" class="field-style field-split25 align-left" placeholder="Name of Vessel" />
但在这种情况下,我得到了Id = ' + 'NameOfVessel' + noOfClicks + '
。据我了解,我需要通过JavaScript来进行连接,只是我无法正确实现。
所以 - 你的mysqli链接/连接可能有问题吗?当你var_dump($ _ POST ['NameOfVessel'])''时,你会得到什么? –