[英]PHP MySQLi Prepared Statement: Fetch Not Returning Results for Joins
我忘記了我從哪里復制它(我想用戶將它發布在php.net上),但是與包含簡單的INNER JOIN的語句一起使用時,用於獲取已准備好的MySQLi語句和普通MySQLi語句的結果的此函數不會返回任何結果。我嘗試過的一個:
function fetch($result){
$array = array();
if($result instanceof mysqli_stmt){
$result->store_result();
$variables = array();
$data = array();
$meta = $result->result_metadata();
while($field = $meta->fetch_field()){
$variables[] = &$data[$field->name];
}
call_user_func_array(array($result, 'bind_result'), $variables);
$i=0;
while($result->fetch()){
$array[$i] = array();
foreach($data as $k=>$v)
$array[$i][$k] = $v;
$i++;
}
}elseif($result instanceof mysqli_result){
while($row = $result->fetch_assoc())
$array[] = $row;
}
return $array;
}
我一直試圖像這樣使用它(所有這些功能都是$ database類的方法):
function query($str, $values){
if($stmt=$this->sql->prepare($str)){
$types=$this->castArrayAsString($values); //returns a string of $values's types for bind_params
array_unshift($values, $types);
call_user_func_array(array(&$stmt, 'bind_param'), $this->makeValuesReferenced($values));
$stmt->execute();
$result=$this->fetch($stmt); //the function I'm having problems with
if(empty($result)){
$return=true;
}else{
$return=$result;
}
$stmt->close();
}else{
echo $this->sql->error;
$return=false;
}
return $return;
}
這不返回任何內容:
$list=$database->query("SELECT a.field1, b.field2 FROM table1 AS a INNER JOIN table2 AS b ON a.id_table2=b.id WHERE a.someid=?", array($someid));
這工作正常:
$list=$database->query("SELECT field1, field2 FROM table1 WHERE someid=?", array($someid));
如果我從獲取結果中打印出元數據,則表明已選中字段,依此類推,但在fetch()函數中仍未給出最終結果。
任何幫助表示贊賞,
謝謝
編輯這里是全班
<?php
class database{
var $HOST="xx.xx.xx.xxx";
var $USER="xxxxxxxxxxx";
var $PASS='xxxxxxxxxx';
var $DATABASE="my_database";
var $sql;
function database(){
$this->sql=new mysqli($this->HOST, $this->USER, $this->PASS, $this->DATABASE);
if($this->sql->connect_errno){
echo "ERROR - No MySQL connection: ".$mysqli->connect_error;
exit;
}
}
function query($str, $values){
if($stmt=$this->sql->prepare($str)){
$types=$this->castArrayAsString($values);
array_unshift($values, $types);
call_user_func_array(array(&$stmt, 'bind_param'), $this->makeValuesReferenced($values));
$stmt->execute();
$result=$this->fetch($stmt);
if(empty($result)){
$return=true;
}else{
$return=$result;
}
$stmt->close();
}else{
echo $this->sql->error;
$return=false;
}
return $return;
}
function fetch($result){
$array = array();
if($result instanceof mysqli_stmt){
$result->store_result();
$variables = array();
$data = array();
$meta = $result->result_metadata();
while($field = $meta->fetch_field()){
$variables[] = &$data[$field->name];
}
call_user_func_array(array($result, 'bind_result'), $variables);
$i=0;
while($result->fetch()){
$array[$i] = array();
foreach($data as $k=>$v)
$array[$i][$k] = $v;
$i++;
}
}elseif($result instanceof mysqli_result){
while($row = $result->fetch_assoc())
$array[] = $row;
}
return $array;
}
function close(){
$this->sql->close();
}
function castArrayAsString($array){
$types="";
foreach($array as $key => $value) {
if(is_numeric($value)){
if(is_float($value)){
$types.="d";
}else{
$types.="i";
}
}else{
$types.="s";
}
}
return $types;
}
function makeValuesReferenced($arr){
$refs=array();
foreach($arr as $key => $value){
$refs[$key] = &$arr[$key];
}
return $refs;
}
}
$database=new database;
?>
返回0結果集的查詢正在執行“內部聯接”,這意味着源表和目標表必須為查詢中要鏈接的列共享相同的值(在本例中為a.id_table2和b。 ID)。 您可能沒有任何匹配值,因此結果集為0。 嘗試左外部聯接。
請嘗試以下操作,看看是否返回結果:
"SELECT a.field1, b.field2 FROM table1 AS a INNER JOIN table2 AS b ON a.id_table2=b.id WHERE a.someid = :someId"
$dbo = new Database();
$stmt = $dbo->sql->prepare($sql);
$stmt->bindValue(':someId', $someId, PDO::PARAM_INT);
$stmt->execute();
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
var_dump($row);
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.