Kik*_*iki 1 php mysql mysqli associative-array fetch
我编写了一个 PHP 脚本,其中返回餐厅名称、地址、电话号码、营业时间表和自定义菜单的链接。但是,即使数据库中有周一的条目,当我在 mysqli_fetch_assoc 中执行 while 循环时,它也不会显示。这是我的代码:
<?php
session_start();
$con=mysqli_connect("root","");
$rest_id2=$_GET['id'];
$rest_id=(int)$rest_id2;
var_dump($rest_id);
$sql="SELECT * FROM restaurant WHERE restaurant_id='".$rest_id."'";
$result=mysqli_query($con,$sql);
$rows=mysqli_fetch_assoc($result);
echo '<strong>'. "Restaurant name:". '</strong><br><br>';
echo $rows['restaurant_name'];
echo "<br><br>";
echo "<strong>Address: </strong><br><br>";
echo $rows['address_1']." ".$rows['address_2']." ". $rows['city']. ", ".
$rows['state']. " ". $rows['zip']. "<br><br>";
echo '<strong>'. "Phone number:". '</strong><br><br>';
echo $rows['phone_number']. "<br><br>";
//hours table
$sql2="SELECT * from hours WHERE restaurant_id='".$rest_id."'";
$result2=mysqli_query($con,$sql2);
$row=mysqli_fetch_assoc($result2);
echo "<table border='1' cellpadding='10'><tr><th>Open or Closed</th> .
<th>Day</th><th>Start Time</th><th>End Time</th></tr>";
$num_rows=mysqli_num_rows($result2);
// var_dump($num_rows);
while($row=mysqli_fetch_assoc($result2)){
// var_dump($rows);
if ($num_rows==0){
echo "No hours data available";
}
elseif($row['day']=="Closed"){
echo "<td><strong>". $row['day']. "</td></strong><br>";
echo "<td><strong>". $row['open_closed']. "</td></strong><br>";
echo "<td><strong>". "-". "</td></strong><br>";
echo "<td><strong>". "-". "</td></tr></strong><br>";
}
else{
echo "<tr><td><strong>". $row['day']. "</td></strong><br>";
echo "<td><strong>". $row['open_closed']. "</td></strong><br>";
echo "<td><strong>". $row['start_time']. "</td></strong><br>";
echo "<td><strong>". $row['end_time']. "</td></tr></strong><br>";
}
}
echo '<a href="' . "custom_menu.php?id=" .$rows['restaurant_id']. '"'.
'>'."<strong>Menu specialized for you</strong>" . '<br>'. '</a>';
?>
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还提供了我在网站上看到的内容。有谁知道为什么会发生这种情况?
小智 5
从代码中删除第 21 行:
$row=mysqli_fetch_assoc($result2);
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它在循环获取其余记录之前获取第一行。while
在循环之外,while您调用第一行并且没有打印它。输出中缺少此行。
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