Question

我尝试使用此查询返回id_selecao：

SELECT id_selecao FROM selecao WHERE nome = 'Brasil'

但没有任何反应，但应该返回id_selecao = 3。但是如果我这样做的话：

SELECT id_selecao FROM selecao WHERE nome = ''

返回id_selecao = 4的更正ID。

这是关于数据库的表格：

id_selecao  nome
1           Portugal
2           Espanha
3           Brasil
4    
5           Portugal
6           Espanha
7           Brasil

继续我的php文件：

initdb.php

<?php
include 'services.php';

function leFicheiroSelecoes(){
    $file_handle = fopen("selecoes.txt", "r");
    while (!feof($file_handle)) {
        $line = fgets($file_handle);

        mysql_set_charset("utf8");
        insertSelecoes($line);
    }
    fclose($file_handle);
}
leFicheiroSelecoes();
disconnect();
?>

services.php

<?php
    /* INCLUDES */
    require('config.php'); //contains the user, pass, dbname to connect and access to database
    require('database.php');

/* DATA BASE */
$db = new database($dbhost, $dbuser, $dbpass, $dbname);



function insertSelecoes($nome){
    global $db;
    $query =  "INSERT INTO selecao (nome) VALUES ('$nome');";
    $db->setQuery($query);
    $db->query();
}

function disconnect(){
    global $db;
    $db->Disconnect();
}
?>

database.php中

<?php
/*
 * Classe que gere a base de dados
 */
class database{

    var $sql = "";
    var $cursor = null;
    var $resource = "";

    function database($host, $user, $password, $db_name) {
        $this->Connect($host, $user, $password, $db_name);
    }

    function Connect($host, $user, $password, $db_name) {
        $this->resource = mysql_connect($host, $user, $password) or die ('Error_connecting_to_my_sql');
        if (!$this->resource) {
            die("Problema na Conex&atilde;o com Base de Dados 1");
        }
        elseif (!mysql_select_db($db_name,$this->resource)) {
            die("Problema na Conex&atilde;o com a Base de Dados 2");
        }
    }

    function Disconnect(){
        return mysql_close($this->resource);
    }

    function query(){
        $this->cursor = mysql_query($this->sql);
        if(!$this->cursor){
            die("Problema em executar a consulta na Base de Dados".mysql_error());
            //return false;
        }
        return $this->cursor;
    }

    function setQuery($sql){
        $this->sql=$sql;
    }

    function loadResult() {
        if (!($result = $this->query())) {
            return null;
        }
        $ret = null;
        if ($row = mysql_fetch_row( $result )) {
            $ret = $row[0];
        }
        mysql_free_result( $result );
        return $ret;
    }

    function loadObjectList( $key='' ) {
        if (!($result = $this->query())) {
            return null;
        }
        $array = array();
        while ($row = mysql_fetch_object( $result )) {
            $array[] = $row;
        }
        mysql_free_result( $result );
        return $array;
    }

}

?>

我无法看到错误在哪里，我一直在寻找这个时间。请有人帮帮我。

Answer 1

你可能在该领域有任何前导或尾随空格？试试这个：

SELECT id_selecao FROM selecao WHERE nome like '%Brasil%'

Answer 2

很可能你在文字'Brasil'周围有一些空白。你可以尝试的是在文本周围添加一些通配符：

SELECT id_selecao FROM selecao WHERE nome like '%Brasil%'

可能会解决这个问题。

Answer 3

$line = fgets($file_handle);

这将返回行，包括最后一行的换行符。来自documentation：

当读取length-1个字节或换行符（包含在返回值中）时，读取结束，...

如果要将其传递给select查询，则应事先切断换行符，例如：

$line = trim ($line);

如果这不起作用，请执行以下操作：

select concat ('[', nome, ']') from selecao where id_selecao = 3

检查数据库表中是否有任何额外的东西。

选择数据库后没有返回任何内容

3 个答案: