Fetch multiple records from php/mysql

Question

I am fetching records as follows:

$aResult = array();
$sql     = "SELECT notes FROM table WHERE user_id = '$user_id'";
$result  = $conn->query($sql);

while($row = mysqli_fetch_array($result)){
    $aResult['query_result'] = $row;
}

This returns only the last record in the table. I need to return all records from the sql statement.

WARNING: When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use string interpolation or concatenation to accomplish this because you have created a severe SQL injection bug. NEVER put $_POST or $_GET data directly into a query, it can be very harmful if someone seeks to exploit your mistake. — tadman
– tadman, Commented Jan 9, 2017 at 5:08

LF-DevJourney · Accepted Answer · 2017-01-09 05:03:25Z

5

change you $aResult['query_result'] = $row; to $aResult['query_result'][] = $row;

You've override the result each time, so you just get one.

answered Jan 9, 2017 at 5:03

LF-DevJourney

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1 Comment

Sanchit Gupta Over a year ago

@KrisRoofe :) Simple and Perfect. +1 Happy Coding.

Paul Coldrey · Accepted Answer · 2017-01-09 05:04:36Z

1

It seems your loop constantly overwrites the value and hence you will only ever seen the last row. I think you might see better results if you do something like:

while($row = mysqli_fetch_array($result))
{
    $aResult[] = $row;
}

so that each new row gets appended to your array

answered Jan 9, 2017 at 5:04

Paul Coldrey

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Comments

Sanchit Gupta · Accepted Answer · 2017-01-09 05:09:50Z

1

Try with following code, currently You are initiating the values to the same array key :

$aResult = array();
$sql     = "SELECT notes FROM table WHERE user_id = '$user_id'";
$result  = $conn->query($sql);

while($row = mysqli_fetch_array($result)){
    $aResult['query_result'][] = $row;
}

for more Detail On Array

answered Jan 9, 2017 at 5:09

Sanchit Gupta

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