Member Avatar for Virangya

hi, i've been trying to solve this but no luck. This is my code

$replace1 =str_replace('hreflang=\"'.$arr['variantslang1hid'].'\" lang=\"'.$arr['variantslang1hid'].'\"','hreflang=\"'.$arr['variantslang1'].'\" lang=\"'.$arr['variantslang1'].'\"',$replace1);

it should replace but no. i'm not escaping "" properly. how can i solve this? help much appreciated!

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  • Latest Post Latest Post by Virangya
Member Avatar for vaultdweller123

try this

$replace1 =str_replace('hreflang="'.$arr['variantslang1hid'].'" lang="'.$arr['variantslang1hid'].'"','hreflang="'.$arr['variantslang1'].'" lang="'.$arr['variantslang1'].'"',$replace1);
Member Avatar for Virangya

try this

$replace1 =str_replace('hreflang="'.$arr['variantslang1hid'].'" lang="'.$arr['variantslang1hid'].'"','hreflang="'.$arr['variantslang1'].'" lang="'.$arr['variantslang1'].'"',$replace1);

thanks for your reply, but i what to escape too..

Member Avatar for Virangya

what?

you have not escaped "" 's n your reply.

Member Avatar for vaultdweller123

i don't get it, why would you escape double quotes(") when it's not even conflicting with your string statement, you wrap them with single quotes, not double quotes that would cause conflict that's why you had to escape them

Member Avatar for Virangya

i don't get it, why would you escape double quotes(") when it's not even conflicting with your string statement, you wrap them with single quotes, not double quotes that would cause conflict that's why you had to escape them

yeah you makes sense. i was the one who messed it up in my head :D. thanks!

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