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我把$query2改成$query2="SELECT `table1` FROM `table2`, `table1` WHERE `$link_id`=$l"; 还是不work....

dream_home(moving2toronto)

说$name=mysql_result($result2, $i, "link_name");这一行

Warning: mysql_result(): supplied argument is not a valid MySQL result resource

source:

----------------------------------
<?

mysql_connect($host,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT * FROM `table1`";

$result=mysql_query($query);

$num=mysql_numrows($result);

$i=0;

while ($i < 15) {

$l=mysql_result($result, $i, "id");

echo $l;

$query2="SELECT `table1` FROM `table2`, `table1` WHERE `$link_id`=$l";

$result2=mysql_query($query2);

$name=mysql_result($result2, $i, "name");

echo ' ';

echo $name;

$i++;
}

mysql_close();
?>

(#2942110@0)
Last Updated: 2006-5-2
This post has been archived. It cannot be replied.

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工作学习 / 专业技术讨论 / 问个php/mysql的问题，同一个裤，两个table, 需要从第一个table里的id 列取一个值(5) 然后找到第2个table里id列(5) 对应的另一个列(name)里的值.......................... 如何？ -dream_home(moving2toronto); 2006-5-2 {526} (#2942019@0)
<?

mysql_connect($host,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT * FROM `table1`";

$result=mysql_query($query);

$num=mysql_numrows($result);

$i=0;

while ($i < 15) {

$l=mysql_result($result, $i, "id");

echo $l;

$query2="SELECT * FROM `table2` WHERE `$link_id`=$l";

$result2=mysql_query($query2);

$name=mysql_result($result2, $i, "name");

echo ' ';

echo $name;

$i++;
}

mysql_close();
?>

use UPDATE -canadiantire(轮胎(do ut des)); 2006-5-2 (#2942090@0)

update will mess up the existing data in tables... i just want to have the results displayed. -dream_home(moving2toronto); 2006-5-2 (#2942095@0)

Sorry, 没仔细看题。呵呵 -canadiantire(轮胎(do ut des)); 2006-5-2 {195} (#2942099@0)
SELECT table2.name FROM table2,table1 WHERE table2.link_id = table1.id

or, if not all link_id map to id, then use

SELECT table2.name FROM table2 LEFT JOIN table1 ON table2.link_id=table1.id

我把$query2改成$query2="SELECT `table1` FROM `table2`, `table1` WHERE `$link_id`=$l"; 还是不work.... -dream_home(moving2toronto); 2006-5-2 {737} (#2942110@0)
说$name=mysql_result($result2, $i, "link_name");这一行

Warning: mysql_result(): supplied argument is not a valid MySQL result resource

source:

----------------------------------
<?

mysql_connect($host,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT * FROM `table1`";

$result=mysql_query($query);

$num=mysql_numrows($result);

$i=0;

while ($i < 15) {

$l=mysql_result($result, $i, "id");

echo $l;

$query2="SELECT `table1` FROM `table2`, `table1` WHERE `$link_id`=$l";

$result2=mysql_query($query2);

$name=mysql_result($result2, $i, "name");

echo ' ';

echo $name;

$i++;
}

mysql_close();
?>

try this one, got to move my car, be back later... -canadiantire(轮胎(do ut des)); 2006-5-2 {404} (#2942142@0)
<?

mysql_connect($host,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT table2.name FROM table1, table2 WHERE table2.link_id=table1.id AND table1.id<15";

$result=mysql_query($query);

$num=mysql_numrows($result);

for($i=0; $i<$num; $i++) {

$name=mysql_result($result, $i);

echo $name." ";

}

mysql_close();
?>

thanks!! -dream_home(moving2toronto); 2006-5-2 {465} (#2942175@0)
thanks!! 还是同样的结果... 我单独query tb2, 也是这样的结果, 难道tb2有问题？对数据类型有要求？

第一个表的第一列叫id, 第2个表也一样，第一个表的id 是有一个顺序的比如 3, 5 ,11, 2 ..... 第二个表的id 也是乱的, 比如4, 22,23,325,234,.... 想读出第一个表的前3位(3,5,11)找到对应表2 id为3,5,11的 name值.....

但是anyway... 即便我 select * from table2 where id=5 还是同样的错误...

table2 的name Type: varchar Collation :latin1_swedish_ci

这些有关系吗？

OK, how about this -canadiantire(轮胎(do ut des)); 2006-5-2 {97} (#2942194@0)
SELECT table2.name FROM table2 WHERE table2.link_id = (SELECT id FROM table1 ORDER BY id LIMIT 3)

Should we add `` around the table name? such as `table1`? -leoxxll(leoxxll); 2006-5-2 (#2942569@0)

Yes you can but you don't have to unless you use keyword or special characters as field name. -canadiantire(轮胎(do ut des)); 2006-5-2 (#2942575@0)

umm... then another thought is, are the table names case-sensitive? -leoxxll(leoxxll); 2006-5-2 (#2942590@0)

Oh, if you are considering the LZ's post, there is already a final solution (#2942527@0). I discussed with LZ it under the table, sorry about that, :-p -canadiantire(轮胎(do ut des)); 2006-5-2 (#2942639@0)

The final corect one ... -canadiantire(轮胎(do ut des)); 2006-5-2 {698} (#2942527@0)
<?php

$link = mysql_connect($server,$username,$password);

mysql_select_db($database);

$query = "SELECT table2.name, table2.link_id FROM table2 INNER JOIN (SELECT link_id FROM table1 ORDER BY table1.date DESC LIMIT 15) AS table2_alias USING (link_id) ";

$result=mysql_query($query, $link);

if( $result ) {

$num=mysql_numrows($result);

for($i=0; $i<$num; $i++) {

$name=mysql_result($result, $i, "name");
$link_id=mysql_result($result, $i, "link_id");

echo '<a href="index.php?option=com_mtree&task=viewlink&link_id='.$link_id.'&Itemid=26">'.$name.'</a> ';

}
} else {
echo "Result is not valid ";
}

mysql_close();
?>

感謝輪胎店.... 全部問題解決... everything is running!!! -dream_home(moving2toronto); 2006-5-2 (#2942678@0)

@Calgary

我把$query2改成$query2="SELECT `table1` FROM `table2`, `table1` WHERE `$link_id`=$l"; 还是不work....

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