Question

在我尝试进行搜索后，我的服务器上一直收到警告：

警告：mysql_fetch_array（）要求参数1为资源，布尔给出第19行的/ search.php

继承代码，我不太明白为什么它不显示结果。任何帮助将非常感激。

<?php
  if(isset($_POST['submit'])){
  if(isset($_GET['go'])){
  if(preg_match("/^[  a-zA-Z]+/", $_POST['postcode'])){
  $postcode=$_POST['postcode'];
  //connect  to the database
  }
$con = mysql_connect('*****','*****','*****'); //connect to the database
if (!$con) //if cannot connect to the database, terminate the process
{
die('Could not connect: ' . mysql_error());
}
  $mydb=mysql_select_db("dataproj");
  //-query  the database table
  $sql="SELECT  * FROM companies WHERE postcode LIKE '%" . $postcode ."%'";
  //-run  the query against the mysql query function
  $result=mysql_query($sql);
  //-create  while loop and loop through result set
  while($row=mysql_fetch_array($result)){
          $postcode =$row['postcode'];
          $cname=$row['cname'];
  //-display the result of the array
  echo "<ul>\n";
  echo "<li>" . $postcode . " " . $cname .  "</a></li>\n";
  echo "</ul>";
  }
  }
  else{
  echo  "<p>Please enter a search query</p>";
  }
  }
?>

Answer 1

首先不要使用不推荐使用的 mysql API使用PDO或mysqli。

其次在将结果传递给mysql_fetch_array之前检查$ result。你会发现这一点它是错误的，因为查询失败了。请参阅mysql_query文档可能的返回值以及如何处理它们的建议。

<?php

if (isset($_POST['submit'])) {
    if (isset($_GET['go'])) {
        if (preg_match("/^[  a-zA-Z]+/", $_POST['postcode'])) {
            $postcode = $_POST['postcode'];
            //connect  to the database
        }
        $con = mysql_connect('*****', '*****', '*****'); //connect to the database
        if (!$con) { //if cannot connect to the database, terminate the process
            die('Could not connect: ' . mysql_error());
        }
        $mydb = mysql_select_db("dataproj");
        //-query  the database table
        $sql = "SELECT  * FROM companies WHERE postcode LIKE '%" . $postcode . "%'";
        //-run  the query against the mysql query function
        $result = mysql_query($sql);
        //check here
        if ($result === FALSE) {
            die(mysql_error()); // TODO: better error handling
        }
        //-create  while loop and loop through result set
        while ($row = mysql_fetch_array($result)) {
            $postcode = $row['postcode'];
            $cname = $row['cname'];
            //-display the result of the array
            echo "<ul>\n";
            echo "<li>" . $postcode . " " . $cname . "</a></li>\n";
            echo "</ul>";
        }
    } else {
        echo "<p>Please enter a search query</p>";
    }
}
?>

PHP / mySQL结果未显示

1 个答案: