来自数据库的数据只显示一个数据,它不显示
Data from database only shows one data and it does not show
我们的 html/php 代码显示数据库中的名称,但其余部分不显示。虽然我们对每个部分都使用了相同的语法,但是我们将数据名称更改为与数据库列名称相同。
为什么会这样?
这是我们的代码:
<?php
$servername = "localhost";
$username = "";
$password = "1234";
$dbname = "straypaws";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dog_name, Dog_loc, Dog_desc1, Dog_desc2, Dog_info1 FROM dog_info";
$result = $conn->query($sql);
?>
<div class="no-name-C61RwL helveticaneue-regular-normal-black-48px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_name"];
} ?>
</div>
<div class="dog-profile-C61RwL helveticaneue-regular-normal-black-72px">DOG PROFILE</div>
<img class="group-49-C61RwL animate-enter7" src="img/group-49@2x.svg" />
<div class="group-51-C61RwL">
<div class="group-49-Hjsg7h">
<div class="burnham-park-baguio-oKkELk helvetica55roman-regular-normal-black-24px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_loc"];
} ?>
您不能多次获取相同的结果。
只需将结果存储在数组中,然后通过 foreach
显示它
$dogs = [];
while($row = $result->fetch_assoc){
$dogs[] = $row;
}
然后在你身上 html:
foreach($dogs as $dog){
echo $dog['Dog_name'];
}
我们的 html/php 代码显示数据库中的名称,但其余部分不显示。虽然我们对每个部分都使用了相同的语法,但是我们将数据名称更改为与数据库列名称相同。
为什么会这样?
这是我们的代码:
<?php
$servername = "localhost";
$username = "";
$password = "1234";
$dbname = "straypaws";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dog_name, Dog_loc, Dog_desc1, Dog_desc2, Dog_info1 FROM dog_info";
$result = $conn->query($sql);
?>
<div class="no-name-C61RwL helveticaneue-regular-normal-black-48px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_name"];
} ?>
</div>
<div class="dog-profile-C61RwL helveticaneue-regular-normal-black-72px">DOG PROFILE</div>
<img class="group-49-C61RwL animate-enter7" src="img/group-49@2x.svg" />
<div class="group-51-C61RwL">
<div class="group-49-Hjsg7h">
<div class="burnham-park-baguio-oKkELk helvetica55roman-regular-normal-black-24px"><?php
if($row = mysqli_fetch_array($result)) {
echo $row["Dog_loc"];
} ?>
您不能多次获取相同的结果。
只需将结果存储在数组中,然后通过 foreach
显示它$dogs = [];
while($row = $result->fetch_assoc){
$dogs[] = $row;
}
然后在你身上 html:
foreach($dogs as $dog){
echo $dog['Dog_name'];
}