如何解决这个问题,我试图用pdo方法从模态形式向mysql数据库插入数据,但那里有些不对劲,我无法找到,请帮助....
这是php代码
<?php
$servername = "localhost";
$username ="root";
$password = "";
$dbname ="la_review";
$cardTable = "product";
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
if (isset($_POST['addProductRev'])) {
echo "<script type='text/javascript'>
$(document).ready(function(){
$('#view-modal').modal('show');
});
</script>";
}
if (isset($_POST['addSubmit'])) {
$addNamaP = $_POST['namaP'];
$addHargaP = $_POST['hargaP'];
$addImageP = $_POST['imageP'];
$addPocessorP = $_POST['processorP'];
$addUkuranResolusiP = $_POST['ukuranResolusiP
'];
$addKapasitasP = $_POST['kapasitasP'];
$addKategoriP = $_POST['kategoriP'];
$addDeskripsiP = $_POST['deskripsiP'];
$addInterfaceP = $_POST['interfaceP'];
$addStmt = $conn->query("INSERT INTO product(name, image, price, processor, ukuranResolusi, kapasitasPenyimpanan, interface, kategori, description) VALUES ('$addNamaP','$addImageP','$addHargaP','$processorP','$addUkuranResolusiP','$addKapasitasP','$addInterfaceP','$addKategoriP','$addDeskripsiP')");
}
?>
这是我的模态代码
<div id="view-modal" class="modal fade" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true" style="display: none;">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title">
<i class="glyphicon glyphicon-plus-sign"></i> Add Product Review
</h4>
</div>
<div class="modal-body">
<div id="modal-loader" style=" text-align: center; margin: 20px;">
<form class="row form-signin" action="index.php" method="post">
<div class="form-group">
<input class="form-control" type="text" name="namaP" class="form-control" placeholder="Nama Produk" required>
</div>
<div class="form-group">
<input class="form-control" type="text" name="hargaP" class="form-control" placeholder="Harga Produk" required>
</div>
<div class="form-group">
<input class="form-control" type="text" name="kategoriP" class="form-control" placeholder="kategori Produk" required>
</div>
<div class="form-group">
<label style="float: left;" for="exampleTextarea">Deskripsi Produk</label>
<textarea name="deskripsiP" class="form-control" id="exampleTextarea" rows="3"></textarea>
</div>
<h4 style="font-weight: bold;">Tabel Spesifikasi</h4>
<div class="form-group">
<input class="form-control" type="text" name="processorP" class="form-control" placeholder="Processor">
</div>
<div class="form-group">
<input class="form-control" type="text" name="ukuranResolusiP" class="form-control" placeholder="Ukuran Produk & Resolusi Layar">
</div>
<div class="form-group">
<input class="form-control" type="text" name="kapasitasP" class="form-control" placeholder="Kapasitas Penyimpanan">
</div>
<div class="form-group">
<input class="form-control" type="text" name="interfaceP" class="form-control" placeholder="Interface Produk">
</div>
<div class="form-group">
<input class="form-control" type="text" name="namaP" class="form-control" placeholder="Nama Produk" required>
</div>
<div class="form-group">
<label style="float: left;" for="exampleInputFile">Image</label>
<input type="file" name="imageP" class="form-control-file" id="exampleInputFile" aria-describedby="fileHelp">
<small id="fileHelp" class="form-text text-muted">Format .png only</small>
</div>
<button class="btn btn-info " type="button" value="submit" name="addSubmit">Submit Produk</button></a>
</form>
</div>
<div class="modal-footer">
<div class="row">
<div col-md-4>
<?php
if ($addStmt) {
echo "success";
}else{
echo "fail";
}
?>
</div>
</div>
</div>
</div>
</div>
</div>
<script type='text/javascript'>
$(document).ready(function(){
$('#view-modal').modal('hide');
});
</script>
答案 0 :(得分:0)
尝试在表单中将按钮的类型设置为submit
,如果设置button
值,则需要在jquery中实现函数onclick
并将ajax post请求发送到后端php函数。只有button
类型的按钮才会提交您的表单。愿这有用。