Like Button: Pass Certain Variable from PHP to Jquery when Clicked

Question

I have this simple php code that displays all users' posts with all of them having a unique 'like' button:

$sql = "SELECT * FROM posts";
$result = mysqli_query($con,$sql);

while($row=mysqli_fetch_assoc($result)){
  $post_content = $row['content'];
  $post_id = $row['id'];

  echo $post_content ."<br/>";
  echo "<a href = '#' id = 'likebutton' onclick = 'like_add($post_id )>Like</a>";
}

And my javascript:

function like_add(post_id){
  alert(post_id); //to test if the link is working
}

I got this error however:

ReferenceError: 2dg7c is not defined

Here 2dg7c is an id of a post in which I clicked LIKE.
Did I missed something?

Actually, I should throw a syntax error, right? Something like: SyntaxError: identifier starts immediately after numeric literal. — PeterMader
– PeterMader, Commented Jul 11, 2017 at 12:08
echo "<a href = '#' id = 'likebutton' onclick = 'like_add($post_id )>Like</a>"; in this line $post_id is a string and you are passing without quotes so it will not work. you have to use `\` — Pankaj Makwana
– Pankaj Makwana, Commented Jul 11, 2017 at 12:10
Incidentally, you're not passing to jQuery, you're passing to JavaScript. Please, quit using alert() for troubleshooting., use console.log() instead. — Jay Blanchard
– Jay Blanchard, Commented Jul 11, 2017 at 12:24
it's working for me . i think you have some other problem @Dranreb — JYoThI
– JYoThI, Commented Jul 11, 2017 at 12:30

Death-is-the-real-truth · Accepted Answer · 2017-07-11 12:41:43Z

3

Change link code like this:-

echo "<a href='#' id='likebutton' onclick='like_add(\"{$post_id}\")'>Like</a>";

edited Jul 11, 2017 at 12:41

answered Jul 11, 2017 at 12:07

Death-is-the-real-truth

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4 Comments

Dranreb Over a year ago

I got an error like this bro ` expected expression, got '}'`

Dranreb Over a year ago

On the first one,, I got an illegal character error. The second one was my original code which is not working..

Dranreb Over a year ago

I followed the second one and still got this error : ReferenceError: 2dg7c is not defined ... Can you post a working snippet of this issue sir? Maybe something's just really wrong with my codes

Death-is-the-real-truth Over a year ago

@Dranreb glad to help you :)

JYoThI · Accepted Answer · 2017-07-11 12:35:18Z

3

You need to escape the double quotes using slash (\) like this

echo "<a href ='#' id ='likebutton' onclick='like_add(\"{$post_id}\")'>Like</a>";

edited Jul 11, 2017 at 12:35

answered Jul 11, 2017 at 12:13

JYoThI

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2 Comments

Jay Blanchard Over a year ago

Why should the OP "use like this"? A good answer will always have an explanation of what was done and why it was done in such a manner, not only for the OP but for future visitors to SO.

Jay Blanchard Over a year ago

I really would also remove the spaces around the =, but this works. Surprised at others not testing their code.There is a space on both sides of the =.

Death-is-the-real-truth · Accepted Answer · 2017-07-11 13:09:51Z

3

Always try to pass value in quotes.

echo "<a href = '#' id = 'likebutton' onclick = 'like_add('".$post_id."' )'>Like</a>";

edited Jul 11, 2017 at 13:09

Death-is-the-real-truth

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answered Jul 11, 2017 at 12:09

Syed Daniyal Asif

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1 Comment

Zaphiel Over a year ago

it wil not work. Because onclick event started with using '. And you trying to do this -> onclick='like_add('2dg7c')'

Zaphiel · Accepted Answer · 2017-07-11 12:20:27Z

2

Try: echo "<a href = '#' id = 'likebutton' onclick = 'like_add(\"$post_id\")'>Like</a>";

You should add " before passing the argument to function. In your state, javascript thinks you referencing a variable as argument. But you want to send string.

edited Jul 11, 2017 at 12:20

answered Jul 11, 2017 at 12:08

Zaphiel

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1 Comment

Monty Khanna Over a year ago

Try this : echo "<a href = '#' id = 'likebutton' onclick = 'like_add($post_id )>Like</a>";

Monty Khanna · Accepted Answer · 2017-07-11 12:20:34Z

2

Try this :

echo '<a href = "#" id = "likebutton" onclick = "like_add({$post_id})">Like</a>';

Reason :

There is a problem of concatenation. Php variable is not binding in string. To bind it we can break the long string and concatenate by dot. or use {} or use heredocs.

edited Jul 11, 2017 at 12:20

answered Jul 11, 2017 at 12:16

Monty Khanna

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5 Comments

Jay Blanchard Over a year ago

Do or do not, there is no "try". A good answer will always have an explanation of what was done and why it was done in such a manner, not only for the OP but for future visitors to SO.

Dranreb Over a year ago

Did not worked either man,, gave me an error.. I think it's because of the "

Monty Khanna Over a year ago

@JayBlanchard updating my answer

Jay Blanchard Over a year ago

Your code returns this <a href = "#" id = "likebutton" onclick = "like_add({$post_id})">Like</a>. Please test before posting an answer. The variable does not get parsed.

JYoThI Over a year ago

inside the single quotes everything is consider as a string

Bibhudatta Sahoo · Accepted Answer · 2017-07-11 12:30:36Z

2

You just missed to close single quote here

echo "<a href = '#' id = 'likebutton' onclick = 'like_add($post_id)>Like</a>";

Just replace the above with

echo "<a href = '#' id = 'likebutton' onclick = 'like_add(\"$post_id\")'>Like</a>";

edited Jul 11, 2017 at 12:30

answered Jul 11, 2017 at 12:09

Bibhudatta Sahoo

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1 Comment

Dranreb Over a year ago

I got this error bro = expected expression, got '}'

James Norman · Accepted Answer · 2017-07-11 13:38:53Z

0

This is how you should write

echo "<a href='#' id='likebutton' onclick=like_add('".$post_id."')>Like</a>";

answered Jul 11, 2017 at 13:38

James Norman

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