本人新手:
$sort=mysql_query("select classes_name from classes order by id");
$classes=mysql_fetch_array($sort);
print_r($classes).'<br>';
打印输出数组也只有一条记录,但是数据库中有好几条,这是这么回事?
$sort=mysql_query("select classes_name from classes order by id");
$classes=mysql_fetch_array($sort);
print_r($classes).'<br>';
打印输出数组也只有一条记录,但是数据库中有好几条,这是这么回事?
while($classes=mysql_fetch_array($sort)){
print_r($classes).'<br>';
}
<form id="form_search" name="form1" method="post" action="">
<p>搜索:
<label for="textfield"></label>
<input type="text" name="textfield" id="textfield" />
</p>
<p>关键词:
<label for="select"></label>
<label for="textfield2"></label>
<input type="text" name="textfield2" id="textfield2" />
</p>
<p>类别:
<?php
$sort=mysql_query("select classes_name from classes order by id");
$classes=mysql_fetch_array($sort); echo '<select name="jumpMenu" id="jumpMenu">';
for($i=0;$i<count($classes);$i++){
?>
<option><?php echo $classes[$i];?></option>
</select>
<?php
}
echo '<br>'.$classes.'<br>';
var_dump($classes);
?>
</p>
<p>
<input type="submit" name="button" id="button" value="搜索" />
</p>
</form>
</body>
其实我是要做一个跳转菜单,你们帮我看看应该怎么写?谢谢啦
while($row = mysql_fetch_array($sort)){
array_push($classes,$row);
}
echo '<select name="jumpMenu" id="jumpMenu">';
for($i=0;$i<count($classes);$i++){
$sort=mysql_query("select classes_name from classes order by id");
echo '<select name="array" id="jumpMenu">';
while($classes=mysql_fetch_array($sort)
){
?>
<option><?php echo $classes['classes_name'];?></option>
<?php
}
echo '</select>';
自己搞定了,谢谢各位!