相关疑难解决方法(0)

如何在不同的环境中获取MySQLi错误信息

在我的本地/开发环境中,MySQLi查询执行正常.但是,当我在我的Web主机环境中上传它时,我收到此错误:

致命错误:在...中的非对象上调用成员函数bind_param()

这是代码:

global $mysqli;
$stmt = $mysqli->prepare("SELECT id, description FROM tbl_page_answer_category WHERE cur_own_id = ?");
$stmt->bind_param('i', $cur_id);
$stmt->execute();
$stmt->bind_result($uid, $desc);
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为了检查我的查询,我试图通过控制面板phpMyAdmin执行查询,结果是OK.

php environment mysqli error-reporting prepared-statement

85
推荐指数
1
解决办法
1万
查看次数

php/mysql有多个查询

<?php

$query1 = "CREATE VIEW current_rankings AS SELECT * FROM main_table WHERE date = X";

$query2 = "CREATE VIEW previous_rankings AS SELECT rank FROM main_table WHERE date = date_sub('X', INTERVAL 1 MONTH)";

$query3 = "CREATE VIEW final_output AS SELECT current_rankings.player, current_rankings.rank as current_rank LEFT JOIN previous_rankings.rank as prev_rank
             ON (current_rankings.player = previous_rankings.player)";

$query4 = "SELECT *, @rank_change = prev_rank - current_rank as rank_change from final_output";

$result = mysql_query($query4) or die(mysql_error()); 

while($row = mysql_fetch_array($result)) {
echo $row['player']. $row['current_rank']. $row['prev_rank']. $row['rank_change'];
} …
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php string concatenation while-loop

7
推荐指数
1
解决办法
5万
查看次数

我在mysql中找不到*确切*语法错误

我正在创建一个简单的数据库,它是用于学习目的的表:

这是我的PHP代码(script.php)

<?php

$sql = file_get_contents("init.sql");

$servername = "localhost";
$username = "root";
$password = "";
  // connect to database
$conn = new mysqli($servername, $username, $password);

if ($conn->connect_error) {
    die("Connection error: " . $conn->connect_error);
}


 if($conn->query($sql) == True){
   echo "Database and Table has been created succesfully!";
 }

 else {
   echo "\nError creating database and table:  . $conn->error";
 }

 ?>
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这是mysql文件(init.mysql)

CREATE DATABASE test;
  USE test;

  CREATE TABLE  Users (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) …
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php mysql

-2
推荐指数
1
解决办法
73
查看次数