PHP for循环 - 在表格中显示数据 [英] PHP for loop - display data in table
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问题描述
我无法理解为什么我的代码不工作;它只显示1个结果,而表中有多个条目。
<?php
$ search = $ _POST ['search'];
print($ search);
$ query =选择*从学生的地方='苏塞克斯';
$ result = mysql_query($ query,$ conn);
$ counter = 0;
$ z = 0;
while($ row = mysql_fetch_assoc($ result)){
$ id [$ counter] = $ row ['roll_number'];
$ counter = $ counter + 1;
}
?>
< table border =1id =table1>
< th> Id< / th>
< th>类型< / th>
< th>说明< / th>
< th>操作员ID< / th>
<?
if($ counter> 0)
{
for($ z = 0; $ z <$ counter; $ z ++)
{
?>
< tr>
<?
if($ z <$ counter)
{
?>
< td>
<?php echo $ id [$ z]?>
< / td>
<?
}
$ z = $ z + 1;
?>
< / tr>
< / table>
解决方案
您忘记关闭FOR循环及其封闭的IF ,我在下面做了:
<?
if($ counter> 0)
{
for($ z = 0; $ z <$ counter; $ z ++)
{
?>
< tr>
<?
if($ z <$ counter)
{
?>
< td>
<?php echo $ id [$ z]?>
< / td>
<?
}
// $ z = $ z + 1; <! - 删除或注释掉此行 - >
?>
< / tr>
<? }
}?> <! - 这里我们关闭for循环,IF - >
< / table>
更新:注意,感谢VinceBurn,$ z增加了两次
i can't understand why my code is not working; it is only displaying 1 result, whereas there are multiple entries in table..
<?php
$search = $_POST['search'];
print ($search);
$query = "SELECT * FROM student WHERE county= 'sussex'";
$result = mysql_query($query,$conn);
$counter = 0;
$z=0;
while($row = mysql_fetch_assoc($result)){
$id[$counter] = $row['roll_number'];
$counter = $counter + 1;
}
?>
<table border="1" id="table1" >
<th>Id</th>
<th>Type</th>
<th>Description</th>
<th>Operator Id</th>
<?
if($counter>0)
{
for($z=0;$z<$counter;$z++)
{
?>
<tr>
<?
if($z<$counter)
{
?>
<td >
<?php echo $id[$z]?>
</td>
<?
}
$z = $z + 1;
?>
</tr>
</table>
解决方案
You forgot to close the FOR loop and its enclosing IF, I've done it below:
<?
if($counter>0)
{
for($z=0;$z<$counter;$z++)
{
?>
<tr>
<?
if($z<$counter)
{
?>
<td >
<?php echo $id[$z]?>
</td>
<?
}
//$z = $z + 1; <!-- REMOVE or comment out this line -->
?>
</tr>
<? }
} ?> <!-- HERE we close the for loop and the IF -->
</table>
UPDATE: noted, thanks to VinceBurn, that $z was incremented twice
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