PHP不将数据库打印为表
我正在尝试将 mySQL 数据库打印到 html 表中。我看过很多关于如何做到这一点的教程，但不确定我如何在我的 php 代码中引用 html 表。信息打印得很好并连接到数据库，但由于某种原因它没有以表格格式输出。
<?php
$conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info');
if (!$conn) {
echo "Connection failed:" . mysqli_connect_error();
}
//Writing query for database.
$sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date Created`";
//Querying and getting results
$result = mysqli_query($conn, $sql);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["First Name"] . "</td></tr>" . $row["Last Name"] . "</td></tr>"
. $row["Emails"] . "</td></tr>" . $row["Date Created"] . "</td></tr>";
}
echo "</table>";
} else {
echo "0 result";
}
//Fetch resulting rows as an array
$informed = mysqli_fetch_all($result, MYSQLI_ASSOC);
// Freeing result from the memory.
mysqli_free_result($result);
mysqli_close($conn);
?>
<!DOCTYPE html>
<html>
<head>
<div class="Contained">
<div class="row">
<?php foreach ($informed as $inform) { ?>
<div class="col s6 medium-3">
<div class="card z-depth-0">
<div class="card-content center">
<h6><?php echo htmlspecialchars($inform['First Name']); ?></h6>
<div><?php echo htmlspecialchars($inform['Last Name']); ?></div>
</div>
<div class="card-action right-align">
<a class="brand-text" href="#">More Info
</div>
</div>
</div>
<?php } ?>
</div>
</div>
<title> Email and Name List </title>
</head>
潇湘沐2回答
-
慕标5832272
您必须更改在 php 中输出所有表的代码，例如：<body><?php$conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info');if (!$conn){echo "Connection failed:" . mysqli_connect_error();}//Writing query for database.$sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date Created`";//Querying and getting results$result = mysqli_query($conn,$sql);if ($result->num_rows>0){echo '<table><tr> <th>First Name</th> <th>Last Name</th> <th>Emails</th> <th>Date Created</th></tr>'; while($row = $result->fetch_assoc()){ echo "<tr> "; echo "<td>" . $row["First Name"] . "</td>"; echo "<td>" . $row["Last Name"] . "</td>"; echo "<td>" . $row["Date Created"] . "</td>"; echo "</tr> "; } echo"</table>";}else{ echo "0 result";}//Fetch resulting rows as an array$informed = mysqli_fetch_all($result, MYSQLI_ASSOC);// Freeing result from the memory.mysqli_free_result($result);mysqli_close($conn);?></body>另一个问题你确定是$row["First Name"]不是$row["First_Name"]？最后提示了解如何准备 stm 以防止 sql 注入 -
墨色风雨
根据您的代码，您尝试在下面定义实际表之前打印该表。你可以尝试这样的事情：<?php $conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info'); if (!$conn){ echo "Connection failed:" . mysqli_connect_error(); } //Writing query for database. $sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date Created`"; //Querying and getting results $result = mysqli_query($conn,$sql); $informed = mysqli_fetch_all($result, MYSQLI_ASSOC); ?><!DOCTYPE html><html><head><div class="Contained"><div class="row"> <?php foreach($informed as $inform){?> <div class="col s6 medium-3"> <div class="card z-depth-0"> <div class="card-content center"> <h6><?php echo htmlspecialchars($inform['First Name']); ?></h6> <div><?php echo htmlspecialchars($inform['Last Name']);?></div> </div> <div class="card-action right-align"> <a class="brand-text" href="#">More Info</div> </div> </div> <?php }?></div></div><title> Email and Name List </title></head><body> <table> <tr> <th>First Name</th> <th>Last Name</th> <th>Emails</th> <th>Date Created</th> </tr> <?php if ($result->num_rows>0) { while($row = $result->fetch_assoc()){ echo "<tr><td>" . $row["First Name"] . "</td><td>" . $row["Last Name"] . "</td><td>" . $row["Emails"] . "</td><td>" . $row["Date Created"] . "</td></tr>"; } } else { echo "<tr><td rowspan=\"5\">0 result</td></tr>"; } ?> </table></body><?php // Freeing result from the memory. mysqli_free_result($result); mysqli_close($conn);?>
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