Question

我正在努力解决这个MySQL错误，我正在使用PHP脚本。

<?php

$connection = mysql_connect('localhost','XXX','XXXX');
mysql_select_db('mybd');

echo "<table>";
echo "<table border = 1>";

echo "</td><td>". "Circuito Estoril" . "</td></tr>";

$query = "SELECT * from tempos GROUP BY piloto";
$result = mysql_query($query);

while($row = mysql_fetch_array($result))
{
    echo "</td><td>" . $row['piloto'];
    foreach($row as $value)
{
    $query = "SELECT * from tempos WHERE piloto = " . $value . "ORDER BY tempo";
    $resultado = mysql_query($query);
    while($rows = mysql_fetch_array($resultado))
    {
        echo "</td><td>" . $rows['tempo'] . "</td><td>" . $rows['data'] . "</td></tr>";
    }

}
}   
echo "</table>";
mysql_close();
?>

我得到的错误是警告：mysql_fetch_array（）期望参数1是资源，第21行的******中给出的布尔值

有人可以指点我吗？

谢谢我提前

Answer 1

将您的查询更改为：

 "SELECT * from `tempos` GROUP BY `piloto`";

和这一个：

$query = "SELECT * from `tempos` WHERE `piloto` = '".$value."'  ORDER BY `tempo`";

在ORDER BY之前放置空格

MySQL错误 - mysql_fetch_array（）期望参数1是资源，给定布尔值

1 个答案: