PHP MySQLi fetch "array push" overrides data

Question

I have 2 arrays:

$arr = [];

$tempArray = [
    'val1' => "xxx",
    'val2' => 0,
    'val3' => 0
];

Then in my mysql query i fill the temp array with values from the current row and finally push him into the $arr:

$stmt->bind_result($tempArray["val1"], $tempArray["val2"], $tempArray["val3"]);

while ( $stmt->fetch () ) {
    array_push($arr, $tempArray);
}

The Problem is, on every loop the "array_push" overrides the data in the $arr.

For example I loop 3 times in the $stmt->fetch().

1. Loop

$tempArray = [
    'val1' => "Hello",
    'val2' => 1,
    'val3' => 2
]

$arr = [
    0 = [
        'val1' => "Hello",
        'val2' => 1,
        'val3' => 2
    ];
]

2. Loop

$tempArray = [
    'val1' => "Stack",
    'val2' => 3,
    'val3' => 4
]

$arr = [
    0 = [
        'val1' => "Stack",
        'val2' => 3,
        'val3' => 4
    ],
    1 = [
        'val1' => "Stack",
        'val2' => 3,
        'val3' => 4
    ];
]

3. Loop

$tempArray = [
    'val1' => "Overflow",
    'val2' => 5,
    'val3' => 6
]

$arr = [
    0 = [
        'val1' => "Overflow",
        'val2' => 5,
        'val3' => 6
    ],
    1 = [
        'val1' => "Overflow",
        'val2' => 5,
        'val3' => 6
    ],
    2 = [
        'val1' => "Overflow",
        'val2' => 5,
        'val3' => 6
    ]
]

I never saw this behavior before and i don't know why it does this.

What i want at the end is this:

$arr = [
    0 = [
        'val1' => "Hello",
        'val2' => 1,
        'val3' => 2
    ],
    1 = [
        'val1' => "Stack",
        'val2' => 3,
        'val3' => 4
    ],
    2 = [
        'val1' => "Overflow",
        'val2' => 5,
        'val3' => 6
    ]
]

$stmt class (requested from @Stevish)

$query = "...";
if ( $stmt = $this->db->prepare($query)) {
        $stmt->bind_param('i',  $xxx);
        $stmt->execute();
        $stmt->store_result();
        $$stmt->bind_result($tempArray["val1"], $tempArray["val2"], $tempArray["val3"]);
        while ( $stmt->fetch () ) {
            $arr[] = $tempArr;
        }
    }

Does this answer your question? bind_result into an array PHP mysqli prepared statement — Dharman
– Dharman ♦, Commented May 5, 2020 at 12:23

alanmanderson · Accepted Answer · 2015-09-19 23:58:21Z

0

The issue is that you are inserting a reference to $tempArray into $arr. Then you change the reference. By the third loop you have 3 references to the same array. That is why the values are showing that way... you can solve this in a rather non intuitive way.

try:

$stmt->bind_result($tempArray["val1"], $tempArray["val2"],$tempArray["val3"]);
while ( $stmt->fetch () ) {
    $x = $tempArray; //This copies the values of $tempArray to $x and each loop will create a new x.
    array_push($arr, $x);
}

answered Sep 19, 2015 at 23:58

alanmanderson

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1 Comment

Dharman Over a year ago

This is never going to work. It doesn't solve the problem with references.

Stevish · Accepted Answer · 2015-09-19 18:09:17Z

0

Try:

while ( $stmt->fetch () ) {
    $arr[] = $tempArray;
}

The [] after the variable simply lets PHP know that you're adding a new entry to the array. It will be given a numerical value (0, 1, 2) as you require.

answered Sep 19, 2015 at 18:09

Stevish

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6 Comments

lhuber Over a year ago

Same behavior as before, data gets overwritten.

Stevish Over a year ago

Ah, I see. I think it's in your $stmt->fetch() function that this is happening. Could you post the code for the $stmt class?

Stevish Over a year ago

Just looked at the man page and saw that your code and mine did the exact same thing, mine just had lower overhead

lhuber Over a year ago

i've added my class in the question above :)

Stevish Over a year ago

Sorry, mate. This is too far outside my area of knowledge. I'm not sure how to work with all that binding, executing and storing. I've not used those commands before. I use an entirely different approach

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