Question

我正在尝试使用一个在for循环中递增的数组来查询我的数据库。出于某种原因，无论我做什么，这都行不通，

我收到错误：

“您的SQL语法出错;请查看与您的MySQL服务器版本对应的手册，以便在第1行附近使用正确的语法”）

当在PHPMyAdmin中查询数据库时，一切都很好，所以我假设for循环中的某些东西搞砸了。有什么想法吗？

我的代码： -

 <?php
$submitclick=$_GET["submitclick"];
if($submitclick==1)
{
require('json.php');
$selec=$_GET["selec"];
$selec=str_replace("'","",$selec);
error_reporting(E_ALL);
mysql_connect("localhost", "seamus", "password") or
    die("Could not connect: " . mysql_error());
mysql_select_db("seamus") or die("No such database");
$sql1 = sprintf("SELECT event FROM attends WHERE student = '%s'",$selec);
$result1 = mysql_query($sql1)
  or die(mysql_error());

while ($row1 = mysql_fetch_array($result1))
 {
  $eve[] = $row1['event'];
 }

for($f=0;$f<count($eve);$f++)
{
$sql = sprintf("SELECT event.id, teaches.staff, day, start, duration, room FROM event JOIN module ON (event.module=module.id) JOIN isin ON (event.id=isin.event) JOIN teaches ON (event.id=teaches.event) JOIN attends ON (event.id=attends.event) WHERE event.id = '%s')",$eve[$f]);
$result = mysql_query($sql)
  or die(mysql_error());  
 $i=0;

while ($row = mysql_fetch_array($result))
    {
     $key =$row['day'].$row['start'];
     $event[$key] = $row['id'];
     $room[$row['id']] = $row['room'];
     $lect[$row['id']] = $row['staff'];
   $time[$i]= $row['duration'];
   $i++;
    }
}     
   $i=0;

Answer 1

此处有一个额外的括号，就在'%s'之后：

... WHERE event.id = '%s')",$eve[$f]);

只需删除它就可以了。

Answer 2

更改行：

$sql = sprintf("SELECT event.id, teaches.staff, day, start, duration, room FROM event JOIN module ON (event.module=module.id) JOIN isin ON (event.id=isin.event) JOIN teaches ON (event.id=teaches.event) JOIN attends ON (event.id=attends.event) WHERE event.id = '%s')",$eve[$f]);

为：

$sql = sprintf("SELECT event.id, teaches.staff, day, start, duration, room FROM event JOIN module ON (event.module=module.id) JOIN isin ON (event.id=isin.event) JOIN teaches ON (event.id=teaches.event) JOIN attends ON (event.id=attends.event) WHERE event.id = %d)",$eve[$f]);

因为event.id是一个整数。

希望有所帮助。

php中的mysql语法错误，输入phpmyadmin时工作正常

2 个答案: