php 资源 ID #4 为什么我会得到这个？
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resource id #4 Why am I getting this?
提问by noob123
I have a pretty basic forum template I am working on for testing purposes
我有一个非常基本的论坛模板，我正在用于测试目的
When I create a topic, and press submit, the proccess updates the database but doesn't output on the screen. Why is this? and Why am I getting a Resource id #4 when I echo the $result from this code below:
当我创建一个主题并按下提交时，过程会更新数据库但不会在屏幕上输出。为什么是这样？当我从下面的代码中回显 $result 时，为什么我会得到一个资源 ID #4：
<?php
$host="server"; // Host name
$username="usernamehere"; // Mysql username
$password=""; // Mysql password
$db_name="forum"; // Database name
$tbl_name="question"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result;
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>
<?php
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
回答by Sibu
You are getting resource id #4because $resultis an resource,you must extract the values contained in it by this way,
你得到的 resource id #4是因为$result是一种资源，你必须通过这种方式提取其中包含的值，
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
更多关于资源变量
Update 2(From OP comments)
更新 2（来自 OP 评论）
You are printing values using field name,In that case you will have to change to
您正在使用字段名称打印值，在这种情况下，您必须更改为
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
或者您可以直接使用mysql_fetch_assoc()，在您的情况下将是
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
回答by ujjwalwahi
Problem is in your code:
问题出在您的代码中：
$result=mysql_query($sql);
echo $result;
$resultis resource type, since mysql_query($sql)returns resource
Stop echoing $result.
$result是资源类型，因为mysql_query($sql)返回资源停止回显$result。
回答by Joddy
If you check the link - http://php.net/manual/en/function.mysql-query.php
如果您检查链接 - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
对于 SELECT、SHOW、DESCRIBE、EXPLAIN 和其他返回结果集的语句，mysql_query() 成功时返回资源，错误时返回 FALSE
Hence you are seeing the Resource#4
因此你看到的是资源#4
. What is it you want to achieve?
. 你想达到什么目的？
回答by Mustaqeem Khan
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
只需检查上面的代码，我认为您的问题应该解决
回答by oxy_moron
You don't have to use mysql_fetch_array(). If you want, try something like this:
您不必使用 mysql_fetch_array()。如果您愿意，请尝试以下操作：
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
