使用PHP从链接的MySQL表输出表
我目前正在尝试用PHP输出MySQL表的结果。我对基于web的编程有基本的了解,但还不足以调试我的代码。我知道SQL很好,数据库链接到我的站点,只需将其发布到表中即可。我将发布代码,并希望得到一些帮助:使用PHP从链接的MySQL表输出表,php,mysql,database,Php,Mysql,Database,我目前正在尝试用PHP输出MySQL表的结果。我对基于web的编程有基本的了解,但还不足以调试我的代码。我知道SQL很好,数据库链接到我的站点,只需将其发布到表中即可。我将发布代码,并希望得到一些帮助: <?php $sql = "SELECT player_name AS 'Name', position AS 'Position', team AS 'Team', opp AS 'Opponent' F
<?php
$sql = "SELECT player_name AS 'Name',
position AS 'Position',
team AS 'Team',
opp AS 'Opponent'
FROM `dbname`
WHERE position = 'QB'";
$stmt = $db->query($sql);
if($stmt-> num_rows > 0) {
echo "<table class='table'>";
echo "<thead class='thead-inverse'>";
echo "<tr><th>Name</th><th>Position</th><th>Team</th><th>Opponent</th>";
echo "</thead>";
echo "<tbody>";
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
echo "<tr><td>";
echo $row['Name'];
echo "</td><td>";
echo $row['Position'];
echo "</td><td>";
echo $row['Team'];
echo "</td><td>";
echo $row['Opponent'];
echo "</td></tr>";
}
echo "</tbody>";
echo "</table>";
}
else {
echo "No Results";
}
您忘记用双引号关闭查询
<?php
$sql = "SELECT player_name AS 'Name',
position AS 'Position',
team AS 'Team',
opp AS 'Opponent'
FROM `dbname`
WHERE position = 'QB'";
$stmt = $db->query($sql);
$stmt -> execute();
if($stmt-> num_rows > 0) {
echo "<table class='table'>";
echo "<thead class='thead-inverse'>";
echo "<tr><th>Name</th><th>Position</th><th>Team</th><th>Opponent</th>";
echo "</thead>";
echo "<tbody>";
while($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
echo "<tr><td>";
echo $row['Name'];
echo "</td><td>";
echo $row['Position'];
echo "</td><td>";
echo $row['Team'];
echo "</td><td>";
echo $row['Opponent'];
echo "</td></tr>";
}
echo "</tbody>";
echo "</table>";
}
else {
echo "No Results";
}
我缩短了查询以便于键入,这是一个输入错误。在实际代码中,我确实用引号结束了查询。问题不在于查询,而在于将结果放入表中。