为什么这个查询只显示一个结果? [英] Why does this query only show one result?
本文介绍了为什么这个查询只显示一个结果?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
下面的查询将使用搜索脚本.出于某种原因,它不会返回任一条件为真的所有结果.我做错了什么?
The query below will be used a search script. For some reason it won't return all results where either condition is true. What am i doing wrong?
$sql = "SELECT name, id_code from codes WHERE name LIKE '%$q%' OR id_code
LIKE '%$q%'";
$result = mysql_query($sql);
$query = mysql_query($sql) or die ("Error: ".mysql_error());
$num_rows1 = mysql_num_rows($result);
if ($result == "")
{
echo "";
}
echo "";
$rows = mysql_num_rows($result);
if($rows == 0)
{
print("<div id=norequests>No results for <strong>$q</strong></div>");
}
elseif($rows > 0)
{
while($row = mysql_fetch_array($query))
{
$name = htmlspecialchars($row['name']);
$code = htmlspecialchars($row['id_code']);
}
print("$code: $name<br /> <br />");
}
}
else{
echo '<div id="error">No results for $q.</div>';
}
推荐答案
您正在 while
之外打印.这意味着,无论您有多少结果,都只会打印一个.
You are printing outside of while
. Which means, no matter how many results you have, only the one will be printed.
要么在循环内打印
while($row = mysql_fetch_array($query))
{
$name = htmlspecialchars($row['name']);
$code = htmlspecialchars($row['id_code']);
print("$code: $name<br /> <br />");
}
或者在循环时收集数组中的变量并在循环后随意使用它们
or collect the variables in an array while looping and use them after the loop as you like
$result_array = array();
while($row = mysql_fetch_array($query))
{
$name = htmlspecialchars($row['name']);
$code = htmlspecialchars($row['id_code']);
$result_array[] = array(
'name' => $name,
'code' => $code
);
}
print_r($result_array);
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