该表Users包含数据但仍显示Records Not Found
<?php
$conn = mysql_connect("localhost", "root", "pass", "Assign1");
$records = mysql_query($conn, "select * from Users");
if(!$records)
{
echo "No Records Found";
exit();
}
while($row = mysql_fetch_array($records))
{
echo $row['name'] . " " . $row['pwd'];
echo "<br />";
}
mysql_close($conn);
?>
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Gli*_*ire 11
你有mysql_query相反的参数.它应该是:
$records = mysql_query("select * from Users", $conn);
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你的另一个问题是if声明.您正在检查if查询,而不是结果集.
此外,我相信您可能知道但是mysql库已被弃用并且正在删除.您应该学会使用mysqli函数,因为它们将来会对您有用.
链接到MySQLi文档 - 它实际上并不比mysql库更难.
要在正确的库中重新实现:
<?php
$mysqli = new mysqli("localhost", "user", "pass", "database");
$query = $mysqli->query("SELECT * FROM users");
$results = $query->fetch_assoc();
if($results) {
foreach($results as $row) {
echo $row['name'] . " " . $row['pwd'] . "<br/>";
}
} else {
echo "No results found.";
}
?>
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希望我不只是为你完成你的整个任务,但是让一个人正确地使用mysqli可能是值得的.
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