php - 更新mysql表中的数据

我正在尝试查看表searchedWords中是否已经存在keyWord。如果是这样，则计数器加一。如果表中不存在它，那么我使用了INSERT。问题是，传递给站点的关键字未存储在数据库中。 BIG的另一个大问题是，计数器未添加。是因为if语句吗？还是while循环？

<?php
date_default_timezone_set('Asia/Manila');
$today = date('m-d-Y');
echo $today;
$urltopost = "http://opac.usls.edu.ph/TLCScripts/interpac.dll?Search";
$datatopost = "FormId=0&Config=pac&LimitsId=0&StartIndex=0&SearchField=7&SearchType=1&ItemsPerPage=20&SearchData=$_POST[keyWord]";
$ch = curl_init ($urltopost);
curl_setopt ($ch, CURLOPT_POST, 1);
curl_setopt ($ch, CURLOPT_POSTFIELDS, $datatopost);
curl_setopt ($ch, CURLOPT_HEADER, 0);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
$returndata = curl_exec ($ch);
echo $returndata;
$con=mysqli_connect("...","...","...","...")or die ('Error: ' . mysql_error());
$sql= "SELECT * FROM searchedWords";
$result= mysqli_query($con,$sql);


 while($row=mysqli_fetch_array($result, MYSQLI_ASSOC))
{
        if($row['keyWord']==$_POST[keyWord])
         {
         $upD="UPDATE searchedWords SET countr = countr + 1 WHERE keyWord = '".$row['keyWord']."'";
          while (!mysqli_query($con,$upD))
          {
           die('Error: ' . mysqli_error($con));
           }
         }
        else
        {
         $insertIn="INSERT INTO `searchedWords`( `keyWord`, `countr`) values ('$_POST[keyWord]',1)";
      while (!mysqli_query($con,$insertIn))
          {
          die('Error: ' . mysqli_error($con));
          }
        }
}
?>

谢谢那些能帮助我的人。

最佳答案

您为什么要使事情变得复杂...尝试简单的方法

$KeyWord = $_POST['keyWord']; //Do not forget to sanitize this for security
$sql= "SELECT * FROM searchedWords WHERE keyWord='$KeyWord'";
$result= mysqli_query($con,$sql);
$count= mysqli_num_rows($result);

if($count) {
$upD="UPDATE searchedWords SET countr = countr + 1 WHERE keyWord = '$keyWord'";
mysqli_query($con,$upD);
} else {
$insertIn="INSERT INTO `searchedWords`( `keyWord`, `countr`) values ('$keyWord',1)";
mysqli_query($con,$insertIn); }

关于php - 更新mysql表中的数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/18987042/