什么导致“未定义索引”?此代码出错


What causes an "Undefined Index" error with this code?

什么是未定义的索引错误?我没有指定任何索引

返回这个错误。Notice: Undefined index: in C:'wamp'www'Maps_ketchupp'search.php on line 73

这是产生错误的代码:

<?php
// Get the search variable from URL
$var = @$_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable
// rows to return
$limit=10; 
// check for an empty string and display a message.
if ($trimmed == "")
{
    echo "<p>Please enter a search...</p>";
    exit;
}
// check for a search parameter
if (!isset($var))
{
    echo "<p>We dont seem to have a search parameter!</p>";
    exit;
}
//connect to your database ** EDIT REQUIRED HERE **
$link = mysql_connect('localhost','root','root'); //(host, username, password)
//specify database ** EDIT REQUIRED HERE **
mysql_select_db("google maps 1.01") or die("Unable to select database"); 
//select which database we're using
// Build SQL Query  
$query = "select * from markers where name like '"%$trimmed%'"  
    order by name"; 
    // EDIT HERE and specify your table and field names for the SQL query
    //$numresults=mysql_query($query);
    //$numrows=mysql_num_rows($numresults);
$numrows=2;
// If we have no results, offer a google search as an alternative
if ($numrows == 2)
{
    echo "<h4>Results</h4>";
    echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>";
    // google
    echo "<p><a href='"http://www.google.com/search?q=" 
        . $trimmed . "'" target='"_blank'" title='"Look up" 
        . $trimmed . " on Google'">Click here</a> to try the 
        search on google</p>";
    }
    // next determine if s has been passed to script, if not use 0
    if (empty($s)) {
        $s=0;
    }
    // get results
    $query .= " limit $s,$limit";
    $result = mysql_query($query) or die("Couldn't execute query");
    // display what the person searched for
    echo "<p>You searched for: &quot;" . $var . "&quot;</p>";
    // begin to show results set
    echo "Results";
    $count = 1 + $s ;
    // now you can display the results returned
    while ($row= mysql_fetch_array($result)) {
        **$title = $row["1st_field"];**-----------SHOWING ERROR IN HIS LINE
        echo "$count.)&nbsp;$title" ;
        $count++ ;
}
$currPage = (($s/$limit) + 1);
//break before paging
echo "<br />";
// next we need to do the links to other results
if ($s>=1) 
{ // bypass PREV link if s is 0
    $prevs=($s-$limit);
    print "&nbsp;<a href='"$PHP_SELF?s=$prevs&q=$var'">&lt;&lt; 
    Prev 10</a>&nbsp&nbsp;";
}
// calculate number of pages needing links
$pages=intval($numrows/$limit);
// $pages now contains int of pages needed unless there is a remainder from division
if ($numrows%$limit) 
{
    // has remainder so add one page
    $pages++;
}
// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1) 
{        
    // not last page so give NEXT link
    $news=$s+$limit;
    echo "&nbsp;<a href='"$PHP_SELF?s=$news&q=$var'">Next 10 &gt;&gt;</a>";
}
$a = $s + ($limit) ;
if ($a > $numrows) { $a = $numrows ; }
$b = $s + 1 ;
echo "<p>Showing results $b to $a of $numrows</p>";
?>

您正在使用mysql_fetch_array而不是mysql_fetch_assoc或您的表没有'1st_field'列

这个错误告诉了你应该怎么做。检查第73行是否有错误。可能是变量名错了,也可能是数据库列名错了。