variable issue for return value - php

Question

trying to return $out
$out = '';
$out .= '<form id="agp_upload_image_form" method="post" action="" enctype="multipart/form-data">';
$out .= wp_nonce_field('agp_upload_image_form', 'agp_upload_image_form_submitted');
$out .= $posttile = get_option('posttitle');
$out .= $postdiscription = get_option('postdiscription');
$out .= $postauthor = get_option('postauthor');
$out .= $postcategory= get_option('postcategory');
$out .= $uploadimage= get_option('uploadimage');
$out .= $posttitleenabledisables = get_option('posttitleenabledisables');
$out .= $postdiscriptionenabledisable = get_option('postdiscriptionenabledisable');
$out .= $postauthorenabledisable = get_option('postauthorenabledisable');
$out .= $postcategoryenabledisable = get_option('postcategoryenabledisable');
$out .= $uploadimageenabledisable = get_option('uploadimageenabledisable');
$out .= $posttaxonomies = get_option('posttaxonomies');
$out .= $enablecaptcha = get_option('captchaprivatekey');
if ($posttitleenabledisables == 'disable') { } else {
$out .= '<label id="labels" for="agp_image_caption">"'.if ( isset($posttile[0])) { echo get_option('posttitle'); } else { echo 'Post Title'; } .'":</label><br/>';
$out .='<input type="text" id="agp_image_caption" name="agp_image_caption" value="$agp_image_caption ;"/><br/>';
}
but stuck at this point giving error
$out .= '<label id="labels" for="agp_image_caption">"'.if ( isset($posttile[0])) { echo get_option('posttitle'); } else { echo 'Post Title'; } .'":</label><br/>';
i want this return but getting error because i think variable not allow if else
can some one tell how to fix this

Answer 1

The way you're trying to use if/else, I assume you need a ternary operator. If/else control cannot be used the way you are trying.
Try:
$out .= '<label id="labels" for="agp_image_caption">"'. ( isset($posttile[0]) ? get_option('posttitle') : 'Post Title' ) .'":</label><br/>';
Basically, if-else does not return any value and cannot be used inline. The ternary operator evaluates to a single value and can be concatenated with strings and used inline in any other expression.
$cond ? $true_val : $false_val
If $cond evaluates to true, the entire statement evaluates to $true_val, otherwise $false_val.

Answer 2

Ty Ternaryoperator.
$out .= '<label id="labels" for="agp_image_caption">"';
$out.= (isset($posttile[0]))? get_option('posttitle') : 'Post Title';
$out.='":</label><br/>';

Answer 3

This is because you have a condition in the string concat.
replace
$out .= '<label id="labels" for="agp_image_caption">"'.if ( isset($posttile[0])) { echo get_option('posttitle'); } else { echo 'Post Title'; } .'":</label><br/>';
with
$caption = isset($posttile[0]) ? get_option('posttitle') : "Post Title";
$out = '<label id="labels" for="agp_image_caption">"' . $caption . '":</label><br/>';

Answer 4

Try like
if ( isset($posttile[0])) {
$out .= '<label id="labels" for="agp_image_caption">'. get_option('posttitle').':</label><br/>';
} else {
$out .= '<label id="labels" for="agp_image_caption">Post Title:</label><br/>';
}
In your code remove extra "
Orelse as like your code change like
$out .= '<label id="labels" for="agp_image_caption">'.if ( isset($posttile[0])) { get_option('posttitle') } else { 'Post Title' } .':</label><br/>';