dongyuan1984 2017-07-10 16:53
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INNER JOIN:注意:显示未定义的索引

I have two table temp_bid and post_project. I want records who bid on projects. and for that, I have made a query using INNER JOIN but, getting Undefined Index error. It works in MySQL editor.

Here is my code.

<?php
include 'conn.php';
session_start();
if ($_SESSION['EmailID'] == "") {
    header("location:index.html");
}

$getTempBid = "SELECT * from temp_bid TB INNER JOIN post_project P ON P._id = TB.prj_id " + 
" WHERE TB.bidBy = '" .$_SESSION['EmailID']."'";

$tempBidResult = mysqli_query($conn,$getTempBid);

while($fetchTempBid = mysqli_fetch_row($tempBidResult)) {
    $tempBidId = $fetchTempBid['tempBidId'];
    $tempBidPrjId = $fetchTempBid['prj_id'];
    $tempBidPostBy = $fetchTempBid['postBy'];
    $tempBidBy = $fetchTempBid['bidBy'];
    $tempBidOn = new DateTime($fetchTempBid['bidOn']);
    $tempBidAmount = $fetchTempBid['amount'];
    $tempBidDays = $fetchTempBid['days'];
    $tempBidProposalDetails = $fetchTempBid['proposalDetails'];
    $tempBidStatus = $fetchTempBid['bidStatus'];
    $prjId = $fetchTempBid['_id'];
    $prjTitle = $fetchTempBid['projectTitle'];
    $prjDescriptions = $fetchTempBid['projectDescriptions'];
    $prjSkills = $fetchTempBid['projectRequiredSkill'];
    $prjSkills1 = explode(",", $prjSkills);
    $prjBudget = $fetchTempBid['projectBudget'];
    $prjPostDate = new DateTime($fetchTempBid['projectCreatedOn']);
    $prjStatus = $fetchTempBid['projectStatus'];
}
$records = mysqli_num_rows($tempBidResult);
?>

Output:

Notice: Undefined index: tempBidId in C:\wamp\www\Gopinath Infosystem\testpage.php on line 23
Notice: Undefined index: prj_id in C:\wamp\www\xxx\testpage.php on line 24
Notice: Undefined index: postBy in C:\wamp\www\xxx\testpage.php on line 25
Notice: Undefined index: bidBy in C:\wamp\www\xxx\testpage.php on line 26
Notice: Undefined index: bidOn in C:\wamp\www\xxx\testpage.php on line 27
Notice: Undefined index: amount in C:\wamp\www\xxx\testpage.php on line 28
Notice: Undefined index: days in C:\wamp\www\xxx\testpage.php on line 29
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1条回答 默认 最新

  • duanlu1908 2017-07-10 16:59
    关注

    mysqli_fetch_row returns an array with numeric keys. From the documentation:

    Fetches one row of data from the result set and returns it as an enumerated array, where each column is stored in an array offset starting from 0 (zero). Each subsequent call to this function will return the next row within the result set, or NULL if there are no more rows.

    Use mysqli_fetch_assoc instead to return an array with the keys being the column names.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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