mysql+php问题

kioyamazaki

php新手。

$sql=mysqli_query($con,"INSERT INTO `CandyWeight`(`Description`, `Qty`, `Weight(g)`) SELECT * FROM (SELECT '$txtdes','$txtqty','$txtweg')as tmp
WHERE NOT EXISTS (SELECT `Description`, `Qty`, `Weight(g)`FROM `CandyWeight` WHERE `Description`='$txtdes' and `Qty`='$txtqty' and `Weight(g)`='$txtweg')";

if (!mysqli_query($con,$sql));
  {
  die('Error: ' . mysqli_error($con));
  }

echo "1 record added";

mysqli_close($con);

问题是无论成功insert与否都会出现error。是不是哪里弄错了。

yan13

可以參考這個.
http://www.w3schools.com/php/func_mysqli_query.asp

geekman

1. SELECT '$txtdes','$txtqty','$txtweg'

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你确定这是对的吗？

flashang

看看你写的：

1. $sql=<font color="Red"><b>mysqli_query</b></font>($con,"INSERT INTO `CandyWeight`(`Description`, `Qty`, `Weight(g)`) SELECT * FROM (SELECT '$txtdes','$txtqty','$txtweg')as tmp
   
2. WHERE NOT EXISTS (SELECT `Description`, `Qty`, `Weight(g)`FROM `CandyWeight` WHERE `Description`='$txtdes' and `Qty`='$txtqty' and `Weight(g)`='$txtweg')";
   
3. 
   

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我解读为，做一次，然后把 结果放入 $sql

1. 
   
2. if (!mysqli_query($con,$sql));
   
3.   {
   
4.   die('Error: ' . mysqli_error($con));
   
5.   }
   
6. 
   
7. echo "1 record added";
   
8. 
   
9. mysqli_close($con);

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第二次是做什么？


你可以尝试这么做。。。

$sql = "insert .... ";
$result = myusqli_query( $con, $sql );

if( .... )