我有一个选择查询,该查询从数据库中提取所有数据,但显示方式如下:
First name: Jasmine
Last name: Santiago
Address: 123 Butt street
City: Butt hill
Customer ID: 12
我正试图让它像这样显示:
First name: Last name: Address: City: Customer ID:
Jasmine Santiago 123 Butt street Butt hill 12
with more rows here of just the data not the labels
但是我有不止一行数据...所以我不确定我要去哪里。这是我的查询和回显语句。
if(isset($_POST["get-user-info"])) {
$user_info_sql = "SELECT * from customer_info";
$user_info_result = $conn->query($user_info_sql);
if ($user_info_result->num_rows > 0) {
while($row = mysqli_fetch_assoc($user_info_result)) {
echo "<tr><th>First name: </th>" . $row["First_name"]. "</tr></br>" . "<tr>Last name: " . $row["Last_name"] . "</tr></br>" . "<tr>Address: " . $row["Address"]. "</tr></br> " . "<tr>City: " . $row["City"] . "</tr></br>" . "Customer ID: " . $row["Customer_id"] . "</br>" . "</br></br>";
}
} else {
echo "No results found";
}
}
您可以看到我有点开始做tr / th / td了,但是我迷路了...我想也许我需要回声像......
回声表 回声tr 回声 ...但是后来我迷路了,因为这会再次呼应标签...
答案 0 :(得分:1)
尝试这样的事情:
<?php
if(isset($_POST["get-user-info"])) {
$user_info_sql = "SELECT * from customer_info";
$user_info_result = $conn->query($user_info_sql);
if ($user_info_result->num_rows > 0) {
echo "<tr><th>First name: </th><th>Last name:</th><th>Address:</th><th>City:</th><th>Customer ID:</th></tr>";
while($row = mysqli_fetch_assoc($user_info_result)) {
echo "<tr><td>". $row["First_name"]."</td><td>" . $row["Last_name"] . "</td><td>" . $row["Address"]. "<td></td>" . $row["City"] . "<td></td>" . $row["Customer_id"] . "</td></tr>";
}
} else {
echo "No results found";
}
}
答案 1 :(得分:1)
您应该只显示一次标题:
if ($user_info_result->num_rows > 0) {
// display table header
echo "<table><tr><th>First name: </th><th>Last name: </th><th>Address: </th><th>City: </th><th>Customer ID: </th></tr>";
while($row = mysqli_fetch_assoc($user_info_result)) {
echo "<tr><td>". $row["First_name"]. "</td>" .
"<td>". $row["Last_name"]. "</td>" .
"<td>". $row["Address"]. "</td>" .
"<td>". $row["City"]. "</td>" .
"<td>". $row["Customer_id"]. "</td></tr>";
}
echo "</table>";
}