Question

我正在使用这个PHP代码来学习数据库和PHP，虽然我收到了错误消息：

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/index.php on line 20

首先，SQL查询结果包含我最近修复过的拼写错误，并且我已经在PHPMyAdmin中直接运行SQL查询 - 它可以正常工作。我找不到使用mysql_error()行的任何合理提示。

以下是代码的其余部分：

<!DOCTYPE html>
<html>
<body>

    <?php

$user_name = "root";
$password = "";
$database = "addressbook";
$server = "localhost";

$db_handle = new mysqli($server, $user_name, $password);
$db_found = $db_handle->select_db($database);

    if ($db_found) {

        $SQL = "SELECT * FROM tb_address_book";
        $result = mysql_query($SQL);

        while ($db_field = mysql_fetch_assoc($result)) {
            print $db_field['ID'] . "<BR>";
            print $db_field['First_Name'] . "<BR>";
            print $db_field['Surname'] . "<BR>";
            print $db_field['Address'] . "<BR>";
        }

        $db_handle->close();

    } else {
        print "Database NOT found";
        $db_handle->close();
    }

?>

</body>
</html>

否则

        if ($result) {

    }

返回false

我做错了什么？

Answer 1

你正在使用过时的函数mysql_query。使用mysqli_query，mysqli_fetch_assoc，你的代码看起来应该是这样的

<?php

$user_name = "root";
$password = "";
$database = "addressbook";
$server = "localhost";

$db_handle = new mysqli($server, $user_name, $password);
$db_found = $db_handle->select_db($database);

if ($db_found) {

    $SQL = "SELECT * FROM tb_address_book";
    $result = mysqli_query($db_found,$SQL);

    while ($db_field = mysql_fetch_assoc($result)) {
        print $db_field['ID'] . "<BR>";
        print $db_field['First_Name'] . "<BR>";
        print $db_field['Surname'] . "<BR>";
        print $db_field['Address'] . "<BR>";
    }

    $db_handle->close();

} else {
    print "Database NOT found";
    $db_handle->close();
}

?>

Answer 2

<%#= Preferably, you should initialize '@contact = Contact.new' in your action and form should be unchanged %>
<%= form_for(Contact.new) do |f| %>

这应该有效。如果没有向我报告

Answer 3

您正在将Mysqli与Mysql混合使用。

使用Mysqli方法查询：

$res = $db_handle->query($SQL);

while ($row = $res->fetch_assoc()) {
    // do stuff
}

这里的文档： http://php.net/manual/en/mysqli.quickstart.statements.php

PHP代码返回mysql_fetch_assoc（），期望参数1是资源

3 个答案: