我有一个下拉列表,在选择任何值后它添加了一些字段的新行,示例如下图所示:
我有 insert.php 用于将值插入 MySQL 数据库。但是有一个问题,只有第一行的值被插入到数据库中。
PHP 看起来像:
foreach($_POST['CertificateType'] as $key => $val){
$CertificateType = $val;
$CertificateType = $_POST['CertificateType'][$key];
$STCWCode = $_POST['STCWCode'][$key];
$CertNo = $_POST['CertNo'][$key];
$FromCert = $_POST['FromCert'][$key];
$ToCert = $_POST['ToCert'][$key];
$CertificateType = mysqli_real_escape_string($link, $CertificateType);
$STCWCode = mysqli_real_escape_string($link, $STCWCode);
$CertNo = mysqli_real_escape_string($link, $CertNo);
$FromCert = mysqli_real_escape_string($link, $FromCert);
$ToCert = mysqli_real_escape_string($link, $ToCert);
$sql3 = "INSERT INTO Tbl (
CertificateType
,UserId
,STCWCode
,CertNo
,FromCert
,ToCert
,DateCreated
) VALUES (
'$CertificateType',
'$UserID',
'$STCWCode',
'$CertNo',
'$FromCert',
'$ToCert',
now())";
if(mysqli_query($link, $sql3)){
echo "Resume created successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
HTML 看起来像:
<fieldset class="fieldset-borders">
<legend>4. Licenses & Certificates</legend>
<ul class="header">
<li>
<select id='options' name="CertificateType[]" class="field-style div-format align-left">
<option selected disabled value="0">Select certificates</option>
<option value="1">One</option>
<option value="2">Two</option>
<option value="3">Three</option>
</select>
</li>
</ul>
<ul class="cert" id="cert">
<li>
<ul class="column">
<li><p class="test-label11">Name</p></li>
</ul>
</li>
<li>
<ul class="column">
<li><p class="test-label11">STCW Code</p></li>
</ul>
</li>
<li>
<ul class="column">
<li><p class="test-label11">Cert. No</p></li>
</ul>
</li>
<li>
<ul class="column">
<li><p class="test-label11">Place of Issue</p></li>
</ul>
</li>
<li>
<ul class="column">
<li><p class="test-label11">Date of Issue</p></li>
</ul>
</li>
<li>
<ul class="column">
<li><p class="test-label11">Date of Expire</p></li>
</ul>
</li>
</ul>
<div class="action2" ></div>
</fieldset>
Javascript 代码,您可以在 JS FIDDLE 查看
我创建了 JS FIDDLE检查表格的那部分。您有解决方法吗?
最佳答案
您需要使用这种格式创建表单
<form action="insert.php" method="post">
<ul>
<li>
<input name="CertificateType[0]" type="hidden">
<input name="CertificateType[0]['STCWCode']" type="text">
<input name="CertificateType[0]['CertNo']" type="text">
<input name="CertificateType[0]['PlaceofIssueCert']" type="text">
<input name="CertificateType[0]['FromCert']" type="date">
<input name="CertificateType[0]['ToCert']" type="date">
</li>
<li>
<input name="CertificateType[1]" type="hidden">
<input name="CertificateType[1]['STCWCode']" type="text">
<input name="CertificateType[1]['CertNo']" type="text">
<input name="CertificateType[1]['PlaceofIssueCert']" type="text">
<input name="CertificateType[1]['FromCert']" type="date">
<input name="CertificateType[1]['ToCert']" type="date">
</li>
...
</ul>
</form>
当您提交此格式的表单时,您将在 insert.php 的 $_POST['CertificateType']
中收到 Numeric Key array
array(
"0" => array(
"STCWCode" => "somevalue",
"CertNo" => "somevalue",
"PlaceofIssueCert" => "somevalue",
"FromCert" => "somevalue",
"ToCert" => "somevalue",
),
"1" => array(
"STCWCode" => "somevalue",
"CertNo" => "somevalue",
"PlaceofIssueCert" => "somevalue",
"FromCert" => "somevalue",
"ToCert" => "somevalue",
),
. . .
)
这里每个索引将代表一行。这可以使用 foreach 循环检索,如下所示:
将您的 php 代码更新为
foreach($_POST['CertificateType'] as $val){
$CertificateType = $val;
$CertificateType = $val;
$STCWCode = $val['STCWCode'];
$CertNo = $val['CertNo'];
$FromCert = $val['FromCert'];
$ToCert = $val['ToCert'];
$CertificateType = mysqli_real_escape_string($link, $CertificateType);
$STCWCode = mysqli_real_escape_string($link, $STCWCode);
$CertNo = mysqli_real_escape_string($link, $CertNo);
$FromCert = mysqli_real_escape_string($link, $FromCert);
$ToCert = mysqli_real_escape_string($link, $ToCert);
$sql3 = "INSERT INTO Tbl (
CertificateType
,UserId
,STCWCode
,CertNo
,FromCert
,ToCert
,DateCreated
) VALUES (
'$CertificateType',
'$UserID',
'$STCWCode',
'$CertNo',
'$FromCert',
'$ToCert',
now())";
if(mysqli_query($link, $sql3)){
echo "Resume created successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
关于javascript - 通过 PHP 从 HTML 表单向数据库插入多个选项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35829682/