Question

我试图从我的表中选择行引用和作者并回显它

我的目标是创建一个随机引用生成器并显示实际引用和作者。

我在表格中输入了25个引号，包含3行（ID，引用，作者）

我的代码如下，我不断收到资源ID＃9错误

<?php

mysql_select_db(name of database);

$quotes = "SELECT author AND quote FROM inspirational_quotes ORDER BY RAND() LIMIT 1";

$result = mysql_query($quotes);

WHILE ($row = mysql_fetch_array($result)):
ENDWHILE; 

echo "$result";
?>

请帮助

Answer 1

首先，我认为你想要

<?php

mysql_select_db(name of database);
$quotes = "SELECT author,quote FROM inspirational_quotes ORDER BY RAND() LIMIT 1";
$result = mysql_query($quotes);

WHILE ($row = mysql_fetch_array($result)):
ENDWHILE; 

echo "$result";
?>

但我还有其他建议

预加载所有报价ID

CREATE TABLE quoteID
(
    ndx int not null auto_increment,
    id int not null,
    PRIMARY KEY (ndx)
);
INSERT INTO quoteID (id) SELECT id FROM inspirational_quotes;

现在根据quoteID表中的id

进行选择

SELECT B.author,B.quote FROM quoteID A INNER JOIN inspirational_quotes B
USING (id) WHERE A.ndx = (SELECT CEILING(MAX(ndx) * RAND()) FROM quoteID);

这应该可以扩展，因为@rnd_id的返回值来自一个在quoteID表中没有间隙的id列表。

<?php

mysql_select_db(name of database);
$quotes = "SELECT B.author,B.quote FROM quoteID A INNER JOIN "
. "inspirational_quotes B USING (id) "
. "WHERE A.ndx = (SELECT CEILING(MAX(ndx) * RAND()) FROM quoteID)";

$result = mysql_query($quotes);

$row = mysql_fetch_array($result);

echo "$result";
?>

试一试!!!

Answer 2

你不能将$ result作为字符串

回显

DO

    WHILE ($row = mysql_fetch_array($result)):
     echo $row['author'] . " " . $row['quote'];
ENDWHILE; 


?>

Answer 3

你没有回应正确的变量。

echo $row['author'] . ": " . $row['quote'];

Answer 4

为什么AND只是逗号。

SELECT author, quote FROM inspirational_quotes ORDER BY RAND() LIMIT 1

<强> MySQL Select syntax

Answer 5

我建议您通过PHP随机化结果以提高性能。例如：

$r = mysql_query("SELECT count(*) FROM inspirational_quotes");
$d = mysql_fetch_row($r);
$rand = mt_rand(0,$d[0] - 1); 
$r = mysql_query("SELECT author,quote FROM inspirational_quotes LIMIT $rand, 1");

从MySQL获取随机行并使用PHP显示它

5 个答案: