blocks|key|3423420|text|首先，不要使用mysql_*()库。它已经过时了。但是，如果您必须使用它，那么就不要使用mysql_result。效率很低。一个更好的过程是使用fetch函数获取整个行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423421|$result+=+mysql_query($sql)+or+die(mysql_error());
$row+=+mysql_fetch_assoc($row);
if+(mysql_num_rows($result)+!=+0)+{
++echo+$row['first_name'],+$row['last_name'],+$row['county'];
}|code-block|syntax|javascript|3423422|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

First, don't use the mysql_*() library. It's deprecated and obsolete. But if you have to use it, then DON'T use mysql_result. It's inefficient. A better process is to fetch an entire row using the fetch functions

<pre><code>$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($row);
if (mysql_num_rows($result) != 0) {
 echo $row['first_name'], $row['last_name'], $row['county'];
}
</code></pre>

blocks|key|3423451|text|该代码应回答OP关于其当前代码的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423452|$query+=+mysql_query("
SELECT+`first_name`,+`last_name`,`county`
FROM++`contact`
WHERE+`first_name`+='".+mysql_real_escape_string(trim($_POST['name']))+."'

");
++(mysql_num_rows($query)+!==+0)+?+print+mysql_result($query,0,'first_name')+:+print+'Name+not+found';
?>|code-block|syntax|javascript|3423453|另外，根据您的错误，请确保这一行：|3423454|`last_name`,`county`|3423455|目前不是|3423456|`last_name`.`county`|3423457|注意逗号/句号的差异。|3423458|或者，如果您试图显示名、姓和县，请使用以下命令：|3423459|$query+=+mysql_query("
SELECT+`first_name`,+`last_name`,`county`
FROM++`contact`
WHERE+`first_name`+='".+mysql_real_escape_string(trim($_POST['name']))+."'

");
++(mysql_num_rows($query)+!==+0)+?+print+mysql_result($query,0,'first_name')."+".mysql_result($query,0,'last_name')."+".mysql_result($query,0,'county')+:+print+'Name+not+found';
?>+|3423460|并最终获得更好的性能和支持的代码使用。|3423461|$mysqli+=+new+mysqli("your_host",+"your_user",+"your_password",+"your_database");

/*+check+connection+*/
if+(mysqli_connect_errno())+{
++++printf("Connect+failed:+%25s\n",+mysqli_connect_error());
++++exit();
}

$query+=+"
SELECT+`first_name`,+`last_name`,`county`
FROM++`contact`
WHERE+`first_name`+='".+mysql_real_escape_string(trim($_POST['name']))+."'

";
$result+=+$mysqli->query($query);

while($row+=+$result->fetch_array()){
++++$rows[]+=+$row;
}

if($result->num_rows+===+0){
++++print+'Name+not+found';
}

foreach($rows+as+$row){
++++print+$row['first_name']."+".$row['last_name']."+".$row['county'];
}

/*+free+result+set+*/
$result->close();

/*+close+connection+*/
$mysqli->close();|3423462|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

This code should answer the OP's question in relation to his current code:

<pre><code>$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'

");
 (mysql_num_rows($query) !== 0) ? print mysql_result($query,0,'first_name') : print 'Name not found';
?&gt;
</code></pre>

Also according to your error, make sure that this line:

<pre><code>`last_name`,`county`
</code></pre>

is not currently

<pre><code>`last_name`.`county`
</code></pre>

Notice the comma/period difference.

OR, if you are trying to display first name, last name, and county use this:

<pre><code>$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'

");
 (mysql_num_rows($query) !== 0) ? print mysql_result($query,0,'first_name')." ".mysql_result($query,0,'last_name')." ".mysql_result($query,0,'county') : print 'Name not found';
?&gt; 
</code></pre>

And ultimately for better performance and supported code use

<pre><code>$mysqli = new mysqli("your_host", "your_user", "your_password", "your_database");

/* check connection */
if (mysqli_connect_errno()) {
 printf("Connect failed: %s\n", mysqli_connect_error());
 exit();
}

$query = "
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'

";
$result = $mysqli-&gt;query($query);

while($row = $result-&gt;fetch_array()){
 $rows[] = $row;
}

if($result-&gt;num_rows === 0){
 print 'Name not found';
}

foreach($rows as $row){
 print $row['first_name']." ".$row['last_name']." ".$row['county'];
}

/* free result set */
$result-&gt;close();

/* close connection */
$mysqli-&gt;close();
</code></pre>

blocks|key|3423553|text|您最简单的方法是使用更新版本的MySQL函数-+mysqli是一个非常好的选择。但是，如果您必须依赖旧的函数集，那么请阅读文档；mysql_result返回数据，您应该在代码中自己准备它。此外，我必须同意@MarcB+-使用mysql_fetch_assoc或类似的东西，当oyu只是和真正期待一一行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423554|但是，如果您有时间，可以合并到mysqli或PDO方法--它们更现代(换句话说，它们一直都是面向对象的)，而且更安全。|3423555|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Your easyest way is to use a newer version of the MySQL functions - mysqli is a very good one to go with. But if you have to rely on the old function set, then pleace read the documentation; mysql_result returns the data, and you should prepare it yourself in the code.
Also, I have to agree with @MarcB - use mysql_fetch_assoc or something like that when oyu are only and really expecting one single row.

But if you have the time, merge to the mysqli or PDO methods - they are more modern (in other words they are OOP all the way) and safer.

I want to get the first name, last name and county to display but I am getting errors when I try to concatenate the echo statement here. How can I correct this? 
 

<pre><code>$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'

");
 echo (mysql_num_rows($query) !== 0)? mysql_result($query,0,'first_name'):'Name not found';
?&gt;
</code></pre>

This is the message I receive:

<blockquote>
 Warning: mysql_result(): last_name.county not found in MySQL result index 5 in C:\xampp\htdocs\ajax_Contact\name.php on line 14
</blockquote>

Concatenate an echo statement having mysql_result

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想要显示名字、姓和县，但是当我试图在这里连接回显语句时，我会收到错误。我该怎么纠正呢？$query = mysql_query("SELECT `first_name`, `last_name`,`county`FROM  `contact`WHERE `first_name` ='". mysql_real_esc...

问连接具有mysql_result的echo语句
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