[英]Unable to retrieve data from database
我的數據庫表中有一些數據,當我嘗試獲取它並使用mysqli_fetch_assoc()
顯示並在網頁上顯示它們時,我失敗了,有人可以幫助我知道我在哪里做錯了。
我的桌子(轎車)由列組成:
saloon_sex
沙龍_地圖
轎車價格
沙龍名稱
沙龍號碼
沙龍圖片
沙龍位置
沙龍哦
Saloon_oh1
沙龍_服務
沙龍_menu1
Saloon_menu2
Saloon_menu3
沙龍_照片1
沙龍_照片2
沙龍_photo3
沙龍區
這是代碼
<?php
$connection = mysqli_connect('localhost','root','') or die(mysqli_error($connection));
mysqli_select_db($connection,'cmssite') or die (mysqli_error($connection));
if($connection){
echo('connected to database');
}
if(isset($_POST['submit'])) {
if(isset($_POST['area'])) {
$search_value = $_POST['area'];
$query = mysqli_query( $connection,"select * from saloon where saloon_area LIKE '%$search_value%'");
if(! $query )
{
die('Could not get data: ' . mysqli_error($query));
}
$row=null;
// $row=mysqli_fetch_assoc($query);
while ($row=mysqli_fetch_array($query))
{ echo $row['saloon_name'];}
?>
<div class="row">
<div class="col-md-4">
<img src="images/toni&guysalon.jpg" height="150" width="150"/></a>
</div>
<div class="col-md-5">
<h3 style="font-weight:bold; margin-top:10px;"><?php echo $row['saloon_name']?></h3>
<img src="images/unisex.png" width="15" height="15"> <?php echo $row['saloon_sex']?>
<div class="clearfix" style="height:10px;"></div>
<img src="images/location.png" width="15" height="15"> Opposite Nerus Emporio, Madhapur
<div class="clearfix" style="height:10px;"></div>
<img src="images/rupee.png" width="15" height="15"> 400+ For Haircut
<div class="clearfix" style="height:10px;"></div>
<img src="images/time.png" width="15" height="15"> Mon to Sun - 10:00 AM to 09:30 PM
</div>
</div>
<?php } }
?>
[p1][1]
<h3 style="font-weight:bold; margin-top:10px;"><?php echo $row['saloon_name']?></h3>
<img src="images/unisex.png" width="15" height="15"> <?php echo $row['saloon_sex']?>
更改
<h3 style="font-weight:bold; margin-top:10px;"><?php echo $row['saloon_name'];?></h3>
<img src="images/unisex.png" width="15" height="15"> <?php echo $row['saloon_sex'];?>
如果要使用條件mysqli_fetch_assoc
或mysqli_fetch_array
,則必須將要打印的代碼保存在條件的mysqli_fetch_assoc
內。 你做錯了,因為你只檢索、打印並關閉它。
while ($row=mysqli_fetch_array($query))
{ echo $row['saloon_name'];}
這是我在打字時會一直做的事情。
while ($row = mysqli_fetch_assoc($query)) {
<div class="col-md-5">
<h3 style="font-weight:bold; margin-top:10px;"><?php echo $row['saloon_name']?></h3>
<img src="images/unisex.png" width="15" height="15"> <?php echo $row['saloon_sex']?>
<div class="clearfix" style="height:10px;"></div>
<img src="images/location.png" width="15" height="15"> Opposite Nerus Emporio, Madhapur
<div class="clearfix" style="height:10px;"></div>
<img src="images/rupee.png" width="15" height="15"> 400+ For Haircut
<div class="clearfix" style="height:10px;"></div>
<img src="images/time.png" width="15" height="15"> Mon to Sun - 10:00 AM to 09:30 PM
</div>
<?php
} // closing tag for mysqli_fetch_assoc/array
?>
您可以通過檢索並將其保存到有意義的變量中來按照您的方式進行操作:
while($row = mysqli_fetch_assoc($query)) {
$saloonSex = $row['saloon_sex'];
$saloonPrice = $row['saloon_price'];
$saloonName = $row['saloon_name'];
... etc etc
<div class="col-md-5">
<h3 style="font-weight:bold; margin-top:10px;"><?php echo $saloonName?></h3>
<img src="images/unisex.png" width="15" height="15"> <?php echo $saloonSex?>
.....etc etc
<?php
} // closing tag for mysqli_fetch_assoc/array
?>
希望它能解決你的問題。
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