dreamer1231 2015-02-27 00:36
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SQL与INNER JOIN不同

I have problem in options If I embed DISTINCT in query I have value= "". If I remove DISTINCT I have value= "1" and I need this.

 $query_result = mysqli_query($db, "SELECT DISTINCT C.name FROM Category AS C INNER JOIN MarketProduct AS MP ON C.Category_ID = MP.ID_Category WHERE MP.ID_Market ='$id'");

<select name="ID_Category">
<?php
while($row = mysqli_fetch_array($query_result)){   
?>

<option value="<?php echo $row['ID_Category'] ; ?>"><?php echo $row['name'] ; ?></option>


<?php
}
?>
</select>
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  • drll85318 2015-02-27 00:58
    关注

    Instead of DISTINCT, use GROUP BY. See below:

      $query_result = mysqli_query($db, "SELECT C.name FROM Category AS C INNER JOIN MarketProduct AS MP ON C.Category_ID = MP.ID_Category WHERE MP.ID_Market ='$id' GROUP BY C.name");
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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