我有一个产品表,我想取回所有产品的名称、价格和图像。但是,什么也没有显示(空白)。连接没问题,我检查过。另外,当我尝试 (print $result;) 检查发生了什么时,它给了我错误(可捕获的 fatal error :类 mysqli_result 的对象无法转换为字符串)。
代码:
<?php
require_once('mysqli_connect.php');
$sql = "select name, price, image from Product where stars>3";
$result = $conn->query($sql);
if ($result){
print $result;
while ($row=mysqli_fetch_assoc($result)){
echo '<section class="products">';
echo '<div class="product-card">';
echo '<div class="product-image"><img src="' . $row['image'] .'"></div>';
echo '<div class="product-info">';
echo '<h4>' . $row['name'] . '</h4>';
echo '<h5>' . $row['price'] . '</h5>';
echo '</div></div></section>';
}
} else {
echo "Couldn't issue db query";
echo mysqli_error($conn);
}
// Close connection
mysqli_close($conn);
?>
最佳答案
将$row = mysqli_fetch_assoc($result)
更改为$row = $result->fetch_assoc()
试试这个:
<?php
require_once('mysqli_connect.php');
$sql = "select name, price, image from Product where stars>3";
$result = $conn->query($sql);
if ($result){
print_r($result);
while ($row = $result->fetch_assoc()){
echo '<section class="products">';
echo '<div class="product-card">';
echo '<div class="product-image"><img src="' . $row['image'] .'"></div>';
echo '<div class="product-info">';
echo '<h4>' . $row['name'] . '</h4>';
echo '<h5>' . $row['price'] . '</h5>';
echo '</div></div></section>';
}
} else {
echo "Couldn't issue db query";
echo mysqli_error($conn);
}
// Close connection
mysqli_close($conn);
?>
关于php - 如何在 php 中使用多个 MySQL 行和值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49409837/