Question

我有一个包含ID,FRNZ,MSTR,TES,DEN,PRT列的表：

ID = id

FRNZ =我们的供应商

MSTR = catalouges的2位数代码

TES =要搜索的代码

DEN =对象的名称

PRT =价格

对于每个CODE，有3行（文章：一个适合一个夹克一条裤子）。我想显示所有3行，其中包含搜索代码的所有详细信息。我编写了这段代码，但我收到了一个错误。

Warning: mysqli_query(): Empty query  on line 20

Warning: mysqli_error() expects exactly 1 parameter, 0 given in line 20

我的代码是

<?php
if(isset($_POST['submit'])){
if(isset($_GET['go'])){ 
$TES=$_POST['name']; 
$db=mysqli_connect ("localhost","root","") or die ('I cannot connect  to the database because: ' . mysqli_error());
 $mydb=mysqli_select_db($db,"prices");
 $sql=`SELECT * FROM PRICES WHERE TES LIKE '%" . $TES . "%' `;
 mysqli_real_escape_string($db,$TES);
 $result=mysqli_query($db,$sql) or die(mysqli_error().'<br>'.$sql);
 while($row=mysqli_fetch_array($result) or die(mysqli_error())){ 
          $FRNZ=$row['FRNZ']; 
          $MSTR=$row['MSTR']; 
          $TES=$row['TES']; 
          $DEN=$row['DEN'];
          $PRT=$row['PRT'];
  }
 echo "<ul>\n"; 
  echo "<li>" . "Furnizor :" .$FRNZ . " Mostrar :" . $MSTR ." Cod tesatura :" .$TES . "Denumire :" . $DEN ."Pret :".$PRT . " </a></li>\n"; 
  echo "</ul>"; 

}
}

else{
echo  "<p>Te rog scrie codul de tesatura</p>";}
?>

Answer 1

您需要修复反引号并引用查询。你有 -

$sql=`SELECT * FROM PRICES WHERE TES LIKE "%" . $TES . "%" `;

应该是 -

$sql="SELECT * FROM PRICES WHERE TES LIKE '%" . $TES . "%' ";

您只需要在运行查询时检查一次查询错误，请更改此项 -

while($row=mysqli_fetch_array($result) or die(mysqli_error())){

到此 -

while($row=mysqli_fetch_array($result)){

无论如何都是空查询

1 个答案: