Fix {add $var}content{/add} generates invalid PHP - #353
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When using {add $var}content{/add} without explicit value assignment,
the generated code was invalid: '$var = if(!isset($var)) { ... }'
Fixed by wrapping the entire assignment in setClose, not just the value:
- If variable is NOT set: capture content into variable
- If variable IS set: discard buffer with ob_end_clean()
Added 3 tests to cover:
- {add $v}Hello{/add} basic capture
- {add $v}Value: {$x}{/add} capture with interpolation
- {set $v = "first"}{add $v}second{/add} verify no overwrite
Fixes #348
Closed
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Fix for Issue #348
Problem
When using
{add \$var}content{/add}without explicit value assignment, the generated PHP code was invalid:This caused a parse error: "syntax error, unexpected token 'if'".
Solution
Changed
setCloseinCompiler.phpto wrap the ENTIRE assignment in the IF condition, not just the value part:This correctly:
Tests Added
{add $v}Hello{/add} Result: {$v}- basic capture{add $v}Value: {$x}{/add}- capture with variable interpolation{set $v = "first"}{add $v}second{/add}- verify {add} does NOT overwrite existing variableFixes
Fixes #348