Question

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);

?>

Answer 1

您的参数mysql_query已反转。它应该是：

$records = mysql_query("select * from Users", $conn);

您的另一个问题是if声明。您在查询时检查if，而不是在结果集上检查。

此外，我相信您可能知道，但mysql库已被弃用且正在删除。您应该学会使用mysqli函数，因为它们将来会对您有用。

Link to MySQLi documentation - 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.";
}

?>

希望我不只是为你完成你的整个任务，但是让一个人正确使用mysqli可能是值得的。

Answer 2

您使用了mysql_query功能

像这样使用它：

 <?php
$conn = mysql_connect("localhost", "root", "pass","Assign1");
$result = mysql_query("select * from Users", $conn);
if(!$records)
{
echo "No Records Found";
exit();
}
while($row = mysql_fetch_array($result))
{
  echo $row['name'] . " " . $row['pwd'];
  echo "<br />";
}
mysql_close($conn);
?>

Answer 3

你去吧

 <?php

 $conn = mysql_connect("localhost", "root", "pass", "Assign1");
 mysql_select_db(' ----your-database-here---', $conn ) ;
 $records = mysql_query($conn, "select * from Users");

 if(mysql_num_rows($records) > 0 )
 {
 while($row = mysql_fetch_array($records))
 {
echo $row['name'] . " " . $row['pwd'];
echo "<br />";
}   


 }else
{
echo "No Records Found";
exit();
}



mysql_close($conn);

?>

Answer 4

让我们先解决这个问题。实际显示的错误是没有选择数据库你必须选择需要代码的数据库

mysql_select_db（ “为Assign1”，$ conn）;在

希望这段代码能完全解决你的问题。试试一次.........

 <?php
$conn = mysql_connect("localhost", "root", "pass");
mysql_select_db("Assign1",$conn);
$result = mysql_query("select * from users", $conn);
if(!$result)
{
    echo "No Records Found";
    exit();
}
while($row = mysql_fetch_array($result))
{
  echo $row[0]['name'];
  echo "<br />";
}
mysql_close($conn);
?>

mysql_query不返回数据

4 个答案: