我需要帮助来解决我的问题。我不能让我的选择器显示92次,我为此写了FOR指令,但它只显示一个值为表格数据库的选择器,我该怎么做... 我的代码:
<?php
$con = mysql_connect("localhost","root","sergios.com");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("phptests", $con);
$result = mysql_query("SELECT * FROM Category");
for($i=1;$i<92;++$i){
echo "<select>";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option>" . $line['name'] . "</option>";
}
echo "</select>";
}
?>
答案 0 :(得分:0)
你需要在循环之外选择开始和结束。由于您已经循环了结果,因此您不需要额外的for循环:
$con = mysql_connect("localhost","root","sergios.com");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("phptests", $con);
$result = mysql_query("SELECT * FROM Category");
echo "<select>";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option>" . $line['name'] . "</option>";
}
echo "</select>";
答案 1 :(得分:0)
您必须在for循环之外构建选项
$result = mysql_query("SELECT * FROM Category");
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
$myOptions .= "<option>" . $line['name'] . "</option>";
}
for($i=1;$i<92;++$i){
echo "<select>";
echo $myOptions;
echo "</select>";
}
答案 2 :(得分:0)
我认为你需要显示92个选择框。尝试使用以下代码....
$result = mysql_query("SELECT * FROM Category");
$options = "";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
$options .= "<option>" . $line['name'] . "</option>";
}
for($i=1;$i<92;++$i){
echo "<select>";
echo $options;
echo "</select>";
}