resource id #4 Why am I getting this?

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I have a pretty basic forum template I am working on for testing purposes

When I create a topic, and press submit, the proccess updates the database but doesn\'t out

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5条回答

旧巷少年郎

2020-12-11 20:56

<?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

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盖世英雄少女心

2020-12-11 21:02

If you check the link - 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

Hence you are seeing the Resource#4

. What is it you want to achieve?

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萌比男神i

2020-12-11 21:07
Problem is in your code:
```
$result=mysql_query($sql);
echo $result;
```
$result is resource type, since mysql_query($sql) returns resource Stop echoing $result.
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佛祖请我去吃肉

2020-12-11 21:15
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
```
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
```
More about resource variable

Update 2(From OP comments)

You are printing values using field name,In that case you will have to change to
```
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
```
Or you can directly use mysql_fetch_assoc(),which in your case will be
```
while($rows=mysql_fetch_assoc($result)){
      echo $rows['id'];
}
```
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执念已碎

2020-12-11 21:15

You don't have to use mysql_fetch_array(). If you want, try something like this:

$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>"

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